We use cookies to provide you the best possible experience on our website. (You don’t need us to show you how to do algebra! What’s needed is a simpler, more intuitive approach! A simpler form of the rule states if y – un, then y = nun – 1*u’. Step 1: Rewrite the square root to the power of ½: D(2cot x) = 2cot x (ln 2), Step 2 Differentiate the inner function, which is &= \left[7\left(x^2 + 1 \right)^6 \cdot (2x) \right](3x – 7)^4 + \left(x^2 + 1 \right)^7 \left[4(3x – 7)^3 \cdot (3) \right] \quad \cmark \end{align*} \] Think something like: “The function is some stuff to the power of 3. Show Solution. We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = \sin x.$ Then $f'(u) = e^u,$ and $g'(x) = \cos x.$ Hence \begin{align*} f'(x) &= e^u \cdot \cos x \8px] The chain rule is a rule for differentiating compositions of functions. √ (x4 – 37) equals (x4 – 37) 1/2, which when differentiated (outer function only!) The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} That’s what we’re aiming for. So the derivative is 7 times that same stuff to the 6th power, times the derivative of that stuff.” $\bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}$ Solutions. Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= -2(\text{stuff})^{-3} \cdot \dfrac{d}{dx}(\cos x – \sin x) \$8px] \text{Then}\phantom{f(x)= }\\ \frac{df}{dx} &= 7(\text{stuff})^6 \cdot \left(\frac{d}{dx}(x^2 + 1)\right) \\[8px] Need to review Calculating Derivatives that don’t require the Chain Rule? Example: Find the derivative of . D(3x + 1)2 = 2(3x + 1)2-1 = 2(3x + 1). • Solution 3. dF/dx = dF/dy * dy/dx So the derivative is 3 times that same stuff to the power of 2, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}$ Here’s a foolproof method: Imagine calculating the value of the function for a particular value of $x$ and identify the steps you would take, because you’ll always automatically start with the inner function and work your way out to the outer function. Knowing where to start is half the battle. Learn More at BYJU’S. The following equation for h ' (x) comes from applying the chain rule incorrectly. In this case, the outer function is the sine function. It is often useful to create a visual representation of Equation for the chain rule. By continuing, you agree to their use. Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(… That is _great_ to hear!! In this presentation, both the chain rule and implicit differentiation will Step 2: Differentiate y(1/2) with respect to y. CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Let’s use the first form of the Chain rule above: \bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} The outer function in this example is 2x. Step 4 Rewrite the equation and simplify, if possible. rule d y d x = d y d u d u d x ecomes Rule) d d x f ( g ( x = f 0 ( g ( x )) g 0 ( x ) \outer" function times of function. Doing so will give us: f'(x)=5•12(5x-2)^3x Which, when simplified, will give us: f'(x)=60(5x-2)^3x And that is our final answer. &= 99\left(x^5 + e^x\right)^{98} \cdot \left(5x^4 + e^x\right) \quad \cmark \end{align*}, Solution 2. It’s more traditional to rewrite it as: Click HERE for a real-world example of the chain rule. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function f ∘ g in terms of the derivatives of f and g. &= -\sin(\tan(3x)) \cdot \sec^2 (3x) \cdot 3 \quad \cmark \end{align*}. Differentiate f (x) =(6x2 +7x)4 f ( x) = ( 6 x 2 + 7 x) 4 . D(4x) = 4, Step 3. Please read and accept our website Terms and Privacy Policy to post a comment. Step 3. This imaginary computational process works every time to identify correctly what the inner and outer functions are. \begin{align*} f(x) &= (\text{stuff})^{-2}; \quad \text{stuff} = \cos x – \sin x \12px] The Chain Rule is a big topic, so we have a separate page on problems that require the Chain Rule. Powered by Create your own unique website with customizable templates. Now, we just plug in what we have into the chain rule. Solution to Example 1. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Buy full access now — it’s quick and easy! Get complete access: LOTS of problems with complete, clear solutions; tips & tools; bookmark problems for later review; + MORE! D(tan √x) = sec2 √x, Step 2 Differentiate the inner function, which is = cos(4x)(4). Most problems are average. The derivative of sin is cos, so: The derivative of ex is ex, but you’ll rarely see that simple form of e in calculus. = e5x2 + 7x – 13(10x + 7), Step 4 Rewrite the equation and simplify, if possible. f ' (x) = (df / du) (du / dx) = - 4 sin (u) (5) We now substitute u = 5x - 2 in sin (u) above to obtain. Step 1: Differentiate the outer function. (The outer layer is the square'' and the inner layer is (3 x +1). This video gives the definitions of the hyperbolic functions, a rough graph of three of the hyperbolic functions: y = sinh x, y = cosh x, y = tanh x Definition •In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. Hint : Recall that with Chain Rule problems you need to identify the “ inside ” and “ outside ” functions and then apply the chain rule. Solution 2 (more formal) . The problems below combine the Product rule and the Chain rule, or require using the Chain rule multiple times. chain rule example problems MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on CHAIN RULE with easy and logical explanations. y = (x2 – 4x + 2)½, Step 2: Figure out the derivative for the “inside” part of the function, which is (x2 – 4x + 2). The outer function in this example is “tan.” (Note: Leave the inner function in the equation (√x) but ignore that too for the moment) The derivative of tan x is sec2x, so: For example, imagine computing \left(x^2+1\right)^7 for x=3. Without thinking about it, you would first calculate x^2 + 1 (which equals 3^2 +1 =10), so that’s the inner function, guaranteed. The inner function is the one inside the parentheses: x 2 … Multiplying 4x3 by ½(x4 – 37)(-½) results in 2x3(x4 – 37)(-½), which when worked out is 2x3/(x4 – 37)(-½) or 2x3/√(x4 – 37). Remember that a function raised to an exponent of -1 is equivalent to 1 over the function, and that an exponent of ½ is the same as a square root function. However, the technique can be applied to a wide variety of functions with any outer exponential function (like x32 or x99. The Practically Cheating Calculus Handbook, The Practically Cheating Statistics Handbook, Chain rule examples: Exponential Functions, https://www.calculushowto.com/derivatives/chain-rule-examples/. The chain rule says dy dx = dy du × du dx and so dy dx = −sinu× 2x = −2xsinx2 Example Suppose we want to diﬀerentiate y = cos2 x = (cosx)2. D(5x2 + 7x – 19) = (10x + 7), Step 3. Step 3: Differentiate the inner function. &= \cos(2x) \cdot 2 \quad \cmark \end{align*}, Solution 2. d/dy y(½) = (½) y(-½), Step 3: Differentiate y with respect to x. Example: Differentiate y = (2x + 1) 5 (x 3 – x +1) 4. Step 2: Differentiate the inner function. √x. Tip: This technique can also be applied to outer functions that are square roots. Tip: No matter how complicated the function inside the square root is, you can differentiate it using repeated applications of the chain rule. We have the outer function f(u) = u^7 and the inner function u = g(x) = x^2 +1. Then f'(u) = 7u^6, and g'(x) = 2x. Then \begin{align*} f'(x) &= 7u^6 \cdot 2x \\[8px] With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. √ X + 1 Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} Let’s use the first form of the Chain rule above: \bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} &= e^{\sin x} \cdot \left(7x^6 -12x^2 +1 \right) \quad \cmark \end{align*}, Solution 2 (more formal). About "Chain Rule Examples With Solutions" Chain Rule Examples With Solutions : Here we are going to see how we use chain rule in differentiation. : (x + 1)½ is the outer function and x + 1 is the inner function. For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. Your first 30 minutes with a Chegg tutor is free! Then you would next calculate $10^7,$ and so $(\boxed{\phantom{\cdots}})^7$ is the outer function. = (sec2√x) ((½) X – ½). Get notified when there is new free material. Solution. Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line). The inner function is the one inside the parentheses: x4 -37. 5x2 + 7x – 19. Solution 2 (more formal). Jump down to problems and their solutions. h ' ( x ) = 2 ( ln x ) This diagram can be expanded for functions of more than one variable, as we shall see very shortly. We’re glad to have helped! Practice: Chain rule intro. Instead, you’ll think something like: “The function is a bunch of stuff to the 7th power. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²)². Chain Rule problems or examples with solutions. You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function. There are lots more completely solved example problems below! Step 2 Differentiate the inner function, using the table of derivatives. : ), Thank you. Worked example: Derivative of √(3x²-x) using the chain rule. &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x)\quad \cmark \8px] Chain Rule Practice Problems: Level 01 Chain Rule Practice Problems : Level 02 If 10 men or 12 women take 40 days to complete a piece of work, how long … This calculus video tutorial shows you how to find the derivative of any function using the power rule, quotient rule, chain rule, and product rule. cot x. Need to use the derivative to find the equation of a … Add the constant you dropped back into the equation. &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] Then. The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. D(cot 2)= (-csc2). Note that I’m using D here to indicate taking the derivative. Want access to all of our Calculus problems and solutions? For example, to differentiate We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. Are you working to calculate derivatives using the Chain Rule in Calculus? We have the outer function f(u) = e^u and the inner function u = g(x) = x^7 – 4x^3 + x. Then f'(u) = e^u, and g'(x) = 7x^6 -12x^2 +1. Hence \begin{align*} f'(x) &= e^u \cdot \left(7x^6 -12x^2 +1 \right)\\[8px] Combine the results from Step 1 (sec2 √x) and Step 2 ((½) X – ½). The chain rule states formally that . Consider a composite function whose outer function is f(x) and whose inner function is g(x). The composite function is thus f(g(x)). Its derivative is given by: \[\bbox[yellow,8px]{ \begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}, Alternatively, if we write $y = f(u)$ and $u = g(x),$ then $\bbox[yellow,8px]{\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} }$. Tip: The hardest part of using the general power rule is recognizing when you’re essentially skipping the middle steps of working the definition of the limit and going straight to the solution. equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \8px] D(e5x2 + 7x – 19) = e5x2 + 7x – 19. &= 8\left(3x^2 – 4x + 5\right)^7 \cdot (6x-4) \quad \cmark \end{align*}. (2x – 4) / 2√(x2 – 4x + 2). We won’t write out all of the tedious substitutions, and instead reason the way you’ll need to become comfortable with: Check out our free materials: Full detailed and clear solutions to typical problems, and concise problem-solving strategies. AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this site. Solution: d d x sin( x 2 os( x 2) d d x x 2 =2 x cos( x 2). Step 4: Simplify your work, if possible. We have Free Practice Chain Rule (Arithmetic Aptitude) Questions, Shortcuts and Useful tips. The results are then combined to give the final result as follows: Combine your results from Step 1 (cos(4x)) and Step 2 (4). Note: keep 3x + 1 in the equation. Recall that \dfrac{d}{du}\left(u^n\right) = nu^{n-1}. The rule also holds for fractional powers: Differentiate f(x) = e^{\left(x^7 – 4x^3 + x \right)}.. du / dx = 5 and df / du = - 4 sin u. Solution 1 (quick, the way most people reason). Differentiate the outer function, ignoring the constant. In this example, no simplification is necessary, but it’s more traditional to write the equation like this: Watch the video for a couple of chain rule examples, or read on below: The formal definition of the chain rule: Want to skip the Summary? -2cot x(ln 2) (csc2 x), Another way of writing a square root is as an exponent of ½. In this example, the inner function is 4x. (b) f(x;y) = xy3 + x 2y 2; @f @x = y3 + 2xy2; @f @y = 3xy + 2xy: (c) f(x;y) = x 3y+ ex; @f @x = 3x2y+ ex; @f As put by George F. Simmons: "if a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man." (10x + 7) e5x2 + 7x – 19. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} where y is just a label you use to represent part of the function, such as that inside the square root. ), Solution 2 (more formal). Although it’s tedious to write out each separate function, let’s use an extension of the first form of the Chain rule above, now applied to $f\Bigg(g\Big(h(x)\Big)\Bigg)$: \bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Bigg(g\Big(h(x)\Big)\Bigg) \right]’ &= f’\Bigg(g\Big(h(x)\Big)\Bigg) \cdot g’\Big(h(x)\Big) \cdot h'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the middle function] } \\[5px]&\qquad \times \text{ [derivative of the middle function, evaluated at the inner function]} \\[5px]&\qquad \quad \times \text{ [derivative of the inner function]}\end{align*}} &= \dfrac{1}{2}\dfrac{1}{ \sqrt{x^2+1}} \cdot 2x \quad \cmark \end{align*}, Solution 2 (more formal). Solutions to Examples on Partial Derivatives 1. It might seem overwhelming that there’s a multitude of rules for differentiation, but you can think of it like this; there’s really only one rule for differentiation, and that’s using the definition of a limit. Differentiate the square'' first, leaving (3 x +1) unchanged. = 2(3x + 1) (3). • Solution 1. The key is to look for an inner function and an outer function. A few are somewhat challenging. \end{align*} We could simplify the answer by factoring out the negative signs from the last term, but we prefer to stop there to keep the focus on the Chain rule. Technically, you can figure out a derivative for any function using that definition. Example: Find d d x sin( x 2). \begin{align*} f(x) &= \big[\text{stuff}\big]^3; \quad \text{stuff} = \tan x \12px] Great problems for practicing these rules. : ), What a great site. \left[\left(x^2 + 1 \right)^7 (3x – 7)^4 \right]’ &= \left[ \left(x^2 + 1 \right)^7\right]’ (3x – 7)^4\, + \,\left(x^2 + 1 \right)^7 \left[(3x – 7)^4 \right]’ \\[8px] In this example, the negative sign is inside the second set of parentheses. We have y = u^7 and u = x^2 +1. Then \dfrac{dy}{du} = 7u^6, and \dfrac{du}{dx} = 2x. Hence \begin{align*} \dfrac{dy}{dx} &= 7u^6 \cdot 2x \\[8px] Tip You can also use this rule to differentiate natural and common base 10 logarithms (D(ln x) = (1/x) and D(log x) = (1/x) log e. Multiplied constants add another layer of complexity to differentiating with the chain rule. Let u = 5x - 2 and f (u) = 4 cos u, hence. Example question: What is the derivative of y = √(x2 – 4x + 2)? No other site explains this nice. &= e^{\sin x} \cdot \cos x \quad \cmark \end{align*}, Solution 2 (more formal). This indicates that the function f(x), the inner function, must be calculated before the value of g(x), the outer function, can be found. For an example, let the composite function be y = √(x4 – 37). So the derivative is -2 times that same stuff to the -3 power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}} Step 1 Differentiate the outer function, using the table of derivatives. Let’s first think about the derivative of each term separately. &= 7(x^2 + 1)^6 \cdot (2x) \quad \cmark \end{align*} Note: You’d never actually write “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Chain rule for partial derivatives of functions in several variables. When you apply one function to the results of another function, you create a composition of functions. We’ll solve this two ways. : ), this was really easy to understand good job, Thanks for letting us know. Hyperbolic Functions - The Basics. Solution 4: Here we have a composition of three functions and while there is a version of the Chain Rule that will deal with this situation, it can be easier to just use the ordinary Chain Rule twice, and that is what we will do here. Just ignore it, for now. Sample problem: Differentiate y = 7 tan √x using the chain rule. Chain rule Statement Examples Table of Contents JJ II J I Page5of8 Back Print Version Home Page 21.2.6 Example Find the derivative d dx h cos ex4 i. We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. The derivative of ex is ex, so: (a) f(x;y) = 3x+ 4y; @f @x = 3; @f @y = 4. f'(x2 – 4x + 2)= 2x – 4), Step 3: Rewrite the equation to the form of the general power rule (in other words, write the general power rule out, substituting in your function in the right places). The derivative of x4 – 37 is 4x(4-1) – 0, which is also 4x3. 1. Example 4: Find the derivative of f(x) = ln(sin(x2)). Chain Rule Example #1 Differentiate $f(x) = (x^2 + 1)^7$. We’ll again solve this two ways. Step 1: Write the function as (x2+1)(½). The derivative of 2x is 2x ln 2, so: It is useful when finding the … The general power rule is a special case of the chain rule, used to work power functions of the form y=[u(x)]n. The general power rule states that if y=[u(x)]n], then dy/dx = n[u(x)]n – 1u'(x). We’re glad you found them good for practicing. Example problem: Differentiate the square root function sqrt(x2 + 1). $\bbox[10px,border:2px solid blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)} }$ Even though few people admit it, almost everyone thinks along the lines of the informal approach in the blue boxes above. Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. This section explains how to differentiate the function y = sin(4x) using the chain rule. We won’t write out “stuff” as we did before to use the Chain Rule, and instead will just write down the answer using the same thinking as above: We can view $\left(x^2 + 1 \right)^7$ as $({\text{stuff}})^7$, where $\text{stuff} = x^2 + 1$. &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*} We could of course simplify the result algebraically to $14x(x^2+1)^2,$ but we’re leaving the result as written to emphasize the Chain rule term $2x$ at the end. Let’s use the first form of the Chain rule above: \bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} &= \sec^2(e^x) \cdot e^x \quad \cmark \end{align*}, Now let’s use the Product Rule: \begin{align*} (f g)’ &= \qquad f’ g\qquad\qquad +\qquad\qquad fg’ \\[8px] Solutions. Using the linear properties of the derivative, the chain rule and the double angle formula , we obtain: {y’\left( x \right) }={ {\left( {\cos 2x – 2\sin x} \right)^\prime } } Differentiate f(x) = \left(3x^2 – 4x + 5\right)^8.. f’ = ½ (x2 – 4x + 2)½ – 1(2x – 4) Solution: In this example, we use the Product Rule before using the Chain Rule. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} We have the outer function $f(u) = u^8$ and the inner function $u = g(x) = 3x^2 – 4x + 5.$ Then $f'(u) = 8u^7,$ and $g'(x) = 6x -4.$ Hence \begin{align*} f'(x) &= 8u^7 \cdot (6x – 4) \\[8px] That isn’t much help, unless you’re already very familiar with it. This section shows how to differentiate the function y = 3x + 12 using the chain rule. Note: keep 5x2 + 7x – 19 in the equation. In this example, 2(3x +1) (3) can be simplified to 6(3x + 1). Even though we had to evaluate f′ at g(x)=−2x+5, that didn't make a difference since f′=6 not matter what its input is. Differentiate $f(x) = (\cos x – \sin x)^{-2}.$, Differentiate $f(x) = \left(x^5 + e^x\right)^{99}.$. Step 4 Simplify your work, if possible. We have the outer function $f(u) = \tan u$ and the inner function $u = g(x) = e^x.$ Then $f'(u) = \sec^2 u,$ and $g'(x) = e^x.$ Hence \begin{align*} f'(x) &= \sec^2 u \cdot e^x \\[8px] With some experience, you won’t introduce a new variable like $u = \cdots$ as we did above. Applying This is a way of breaking down a complicated function into simpler parts to differentiate it piece by piece. Combine the results from Step 1 (2cot x) (ln 2) and Step 2 ((-csc2)). —– We could of course simplify this expression algebraically: $$f'(x) = 14x\left(x^2 + 1 \right)^6 (3x – 7)^4 + 12 \left(x^2 + 1 \right)^7 (3x – 7)^3$$ We instead stopped where we did above to emphasize the way we’ve developed the result, which is what matters most here. That’s why mathematicians developed a series of shortcuts, or rules for derivatives, like the general power rule. Identify the mistake(s) in the equation. Combine the results from Step 1 (e5x2 + 7x – 19) and Step 2 (10x + 7). Step 2 Differentiate the inner function, which is We have the outer function $f(u) = u^{99}$ and the inner function $u = g(x) = x^5 + e^x.$ Then $f'(u) = 99u^{98},$ and $g'(x) = 5x^4 + e^x.$ Hence \begin{align*} f'(x) &= 99u^{98} \cdot (5x^4 + e^x) \\[8px] In integration, the counterpart to the chain rule is the substitution rule . = (2cot x (ln 2) (-csc2)x). Include the derivative you figured out in Step 1: : ), Thanks! x(x2 + 1)(-½) = x/sqrt(x2 + 1). Step 2:Differentiate the outer function first. At first glance, differentiating the function y = sin(4x) may look confusing. In this example, the inner function is 3x + 1. We’re happy to have helped! 1. In this case, the outer function is x2. How can I tell what the inner and outer functions are? Just ignore it, for now. Once you’ve performed a few of these differentiations, you’ll get to recognize those functions that use this particular rule. Think something like: “The function is some stuff to the $-2$ power. We have the outer function $f(u) = \sqrt{u}$ and the inner function $u = g(x) = x^2 + 1.$ Then $\left(\sqrt{u} \right)’ = \dfrac{1}{2}\dfrac{1}{ \sqrt{u}},$ and $\left(x^2 + 1 \right)’ = 2x.$ Hence \begin{align*} f'(x) &= \dfrac{1}{2}\dfrac{1}{ \sqrt{u}} \cdot 2x \\[8px] Partial derivative is a method for finding derivatives of multiple variables. Covered for all Bank Exams, Competitive Exams, Interviews and Entrance tests. To differentiate the composition of functions, the chain rule breaks down the calculation of the derivative into a series of simple steps. Step 1: Identify the inner and outer functions. For instance, $\left(x^2+1\right)^7$ is comprised of the inner function $x^2 + 1$ inside the outer function $(\boxed{\phantom{\cdots}})^7.$ As another example, $e^{\sin x}$ is comprised of the inner function $\sin x$ inside the outer function $e^{\boxed{\phantom{\cdots}}}.$ As yet another example, $\ln{(t^3 – 2t^2 +5)}$ is comprised of the inner function $t^3 – 2t^2 +5$ inside the outer function $\ln(\boxed{\phantom{\cdots}}).$ Since each of these functions is comprised of one function inside of another function — known as a composite function — we must use the Chain rule to find its derivative, as shown in the problems below. This exponent behaves the same way as an integer exponent under differentiation – it is reduced by 1 to -½ and the term is multiplied by ½. Step 5 Rewrite the equation and simplify, if possible. 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