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. 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. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. \end{align*}. We’re happy to have helped! Huge thumbs up, Thank you, Hemang! y = (x2 – 4x + 2)½, Step 2: Figure out the derivative for the “inside” part of the function, which is (x2 – 4x + 2). It’s more traditional to rewrite it as: Are you working to calculate derivatives using the Chain Rule in Calculus? We’re glad you found them good for practicing. Differentiate $f(x) = (\cos x – \sin x)^{-2}.$, Differentiate $f(x) = \left(x^5 + e^x\right)^{99}.$. = f’ = ½ (x2-4x + 2) – ½(2x – 4), Step 4: (Optional)Rewrite using algebra: Example problem: Differentiate the square root function sqrt(x2 + 1). The second is more formal. &= 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. Solution 2 (more formal). 5x2 + 7x – 19. : ), Thank you. Solution 1 (quick, the way most people reason). In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . We use cookies to provide you the best possible experience on our website. Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line). Hyperbolic Functions - The Basics. Solution 2 (more formal). Knowing where to start is half the battle. (a) f(x;y) = 3x+ 4y; @f @x = 3; @f @y = 4. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. The Chain Rule is a big topic, so we have a separate page on problems that require the Chain Rule. Find the derivatives equation: az dx2 : az дхду if z(x,y) is given by the ay 23 + 3xyz = 1 3. d/dx sqrt(x) = d/dx x(1/2) = (1/2) x(-½). 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. √ (x4 – 37) equals (x4 – 37) 1/2, which when differentiated (outer function only!) • Solution 1. &= -\sin(\tan(3x)) \cdot \sec^2 (3x) \cdot 3 \quad \cmark \end{align*}. Solution: The derivatives of f and g aref′(x)=6g′(x)=−2.According to the chain rule, h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12. By continuing, you agree to their use. \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= -2(\text{stuff})^{-3} \cdot \dfrac{d}{dx}(\cos x – \sin x) \\[8px] Now, we just plug in what we have into the chain rule. Step 2 Differentiate the inner function, which is (10x + 7) e5x2 + 7x – 19. Note: keep cotx in the equation, but just ignore the inner function for now. Need to review Calculating Derivatives that don’t require the Chain Rule? Chain Rule - Examples. Combine the results from Step 1 (2cot x) (ln 2) and Step 2 ((-csc2)). Hyperbolic Functions And Their Derivatives. So the derivative is 7 times that same stuff to the 6th power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] This is the currently selected item. We have the outer function $f(u) = u^3$ and the inner function $u = g(x) = \tan x.$ Then $f'(u) = 3u^2,$ and $g'(x) = \sec^2 x.$ (Recall that $(\tan x)’ = \sec^2 x.$) Hence \begin{align*} f'(x) &= 3u^2 \cdot (\sec^2 x) \\[8px] Think something like: “The function is some stuff to the power of 3. \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= 3\big[\text{stuff}\big]^2 \cdot \dfrac{d}{dx}(\tan x) \\[8px] Think something like: “The function is some stuff to the $-2$ power. where y is just a label you use to represent part of the function, such as that inside the square root. 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. Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(… Differentiate the outer function, ignoring the constant. In this example, the inner function is 3x + 1. D(cot 2)= (-csc2). &= \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] 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] That’s what we’re aiming for. &= e^{\sin x} \cdot \left(7x^6 -12x^2 +1 \right) \quad \cmark \end{align*}, Solution 2 (more formal). If 30 men can build a wall 56 meters long in 5 days, what length of a similar wall can be built … = e5x2 + 7x – 13(10x + 7), Step 4 Rewrite the equation and simplify, if possible. D(sin(4x)) = cos(4x). How can I tell what the inner and outer functions are? Question 1 : Differentiate f(x) = x / √(7 - 3x) Solution : u = x. u' = 1. v = √(7 - 3x) v' = 1/2 √(7 - 3x)(-3) ==> -3/2 √(7 - 3x)==>-3/2 √(7 - 3x) Step 2: Differentiate the inner function. This 105. is captured by the third of the four branch diagrams on the previous page. Differentiate $f(x) = \left(3x^2 – 4x + 5\right)^8.$. x(x2 + 1)(-½) = x/sqrt(x2 + 1). Use the chain rule to calculate h′(x), where h(x)=f(g(x)). cot x. We have the outer function $f(u) = \sin u$ and the inner function $u = g(x) = 2x.$ Then $f'(u) = \cos u,$ and $g'(x) = 2.$ Hence \begin{align*} f'(x) &= \cos u \cdot 2 \\[8px] D(√x) = (1/2) X-½. The derivative of sin is cos, so: Step 2 Differentiate the inner function, using the table of derivatives. = cos(4x)(4). Step 4: Simplify your work, if possible. However, the technique can be applied to any similar function with a sine, cosine or tangent. : ), What a great site. Use the chain rule to differentiate composite functions like sin(2x+1) or [cos(x)]³. The key is to look for an inner function and an outer function. Step 2:Differentiate the outer function first. Include the derivative you figured out in Step 1: At first glance, differentiating the function y = sin(4x) may look confusing. 1. 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] Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. 2x * (½) y(-½) = x(x2 + 1)(-½), Step 5: Simplify your answer by writing it in terms of square roots. It is often useful to create a visual representation of Equation for the chain rule. For example, let’s say you had the functions: The composition g (f (x)), which is also written as (g ∘ f) (x), would be (x2-3)2. 7 (sec2√x) ((1/2) X – ½). The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. = (sec2√x) ((½) X – ½). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] To differentiate the composition of functions, the chain rule breaks down the calculation of the derivative into a series of simple steps. 7 (sec2√x) ((½) X – ½) = Step 1 Differentiate the outer function, using the table of derivatives. Solution 2 (more formal) . Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] This is called a tree diagram for the chain rule for functions of one variable and it provides a way to remember the formula (Figure \(\PageIndex{1}\)). by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²)². Step 1: Identify the inner and outer functions. Step 2: Differentiate y(1/2) with respect to y. A simpler form of the rule states if y – un, then y = nun – 1*u’. In this example, the negative sign is inside the second set of parentheses. Example: Differentiate y = (2x + 1) 5 (x 3 – x +1) 4. 7 (sec2√x) ((½) 1/X½) = In this example, no simplification is necessary, but it’s more traditional to write the equation like this: Step 5 Rewrite the equation and simplify, if possible. Example problem: Differentiate y = 2cot x using the chain rule. Watch the video for a couple of chain rule examples, or read on below: The formal definition of the chain rule: h ' ( x ) = 2 ( ln x ) Identify the mistake(s) in the equation. 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: Step 3: Differentiate the inner function. (2x – 4) / 2√(x2 – 4x + 2). Step 3: Combine your results from Step 1 2(3x+1) and Step 2 (3). The chain rule states formally that . (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 Solutions to Examples on Partial Derivatives 1. Hint : Recall that with Chain Rule problems you need to identify the “ inside ” and “ outside ” functions and then apply the chain rule. √ X + 1 Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] It is useful when finding the … Powered by Create your own unique website with customizable templates. \begin{align*} f(x) &= \big[\text{stuff}\big]^3; \quad \text{stuff} = \tan x \\[12px] Chain Rule: The General Power Rule The general power rule is a special case of the chain rule. 1. \begin{align*} f(x) &= (\text{stuff})^7; \quad \text{stuff} = x^2 + 1 \\[12px] &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x)\quad \cmark \\[8px] The outer function in this example is 2x. = 2(3x + 1) (3). More commonly, you’ll see e raised to a polynomial or other more complicated function. Thanks for letting us know! Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Once you’ve performed a few of these differentiations, you’ll get to recognize those functions that use this particular rule. A few are somewhat challenging. Great problems for practicing these rules. f ' (x) = - 20 sin (5x - 2) We’ll solve this two ways. 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] \(g\left( t \right) = {\left( {4{t^2} - 3t + 2} \right)^{ - 2}}\) Solution. Since the functions were linear, this example was trivial. In this case, the outer function is x2. Worked example: Derivative of √(3x²-x) using the chain rule. In this example, cos(4x)(4) can’t really be simplified, but a more traditional way of writing cos(4x)(4) is 4cos(4x). &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] du / dx = 5 and df / du = - 4 sin u. &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] Step 4 Rewrite the equation and simplify, if possible. Just ignore it, for now. 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] Let f(x)=6x+3 and g(x)=−2x+5. 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. Step 3. Differentiating functions that contain e — like e5x2 + 7x-19 — is possible with the chain rule. &= \dfrac{1}{2}\dfrac{1}{ \sqrt{x^2+1}} \cdot 2x \quad \cmark \end{align*}, Solution 2 (more formal). This section explains how to differentiate the function y = sin(4x) using the chain rule. So the derivative is $-2$ times that same stuff to the $-3$ power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] The chain rule can be used to differentiate many functions that have a number raised to a power. 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. &= \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*} \] D(tan √x) = sec2 √x, Step 2 Differentiate the inner function, which is dF/dx = dF/dy * dy/dx For an example, let the composite function be y = √(x4 – 37). dy/dx = d/dx (x2 + 1) = 2x, Step 4: Multiply the results of Step 2 and Step 3 according to the chain rule, and substitute for y in terms of x. We now use the chain rule. Example 4: Find the derivative of f(x) = ln(sin(x2)). Let u = 5x - 2 and f (u) = 4 cos u, hence. The following chain rule examples show you how to differentiate (find the derivative of) many functions that have an “ inner function ” and an “ outer function.” For an example, take the function y = √ (x 2 – 3). Example: Find d d x sin( x 2). We’re glad to have helped! We have Free Practice Chain Rule (Arithmetic Aptitude) Questions, Shortcuts and Useful tips. Partial derivative is a method for finding derivatives of multiple variables. Step 1: Write the function as (x2+1)(½). The inner function is the one inside the parentheses: x4 -37. : ), Thanks! • Solution 2. In order to use the chain rule you have to identify an outer function and an inner function. D(4x) = 4, Step 3. In this example, the outer function is ex. 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 Then. Then you would next calculate $10^7,$ and so $(\boxed{\phantom{\cdots}})^7$ is the outer function. Technically, you can figure out a derivative for any function using that definition. equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] D(3x + 1) = 3. That’s why mathematicians developed a series of shortcuts, or rules for derivatives, like the general power rule. Solution. The following equation for h ' (x) comes from applying the chain rule incorrectly. The derivative of 2x is 2x ln 2, so: AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this site. You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function. : ). This section shows how to differentiate the function y = 3x + 12 using the chain rule. Worked example: Derivative of ln(√x) using the chain rule. Show Solution. This diagram can be expanded for functions of more than one variable, as we shall see very shortly. Add the constant you dropped back into the equation. 7 (sec2√x) / 2√x. &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*}. Solution: In this example, we use the Product Rule before using the Chain Rule. Get an idea on partial derivatives-definition, rules and solved examples. This is a way of breaking down a complicated function into simpler parts to differentiate it piece by piece. D(2cot x) = 2cot x (ln 2), Step 2 Differentiate the inner function, which is Solutions. Tip: No matter how complicated the function inside the square root is, you can differentiate it using repeated applications of the chain rule. Differentiating using the chain rule usually involves a little intuition. Step 1 Differentiate the outer function. The derivative of ex is ex, so: Step 1 The derivative of ex is ex, but you’ll rarely see that simple form of e in calculus. The chain rule in calculus is one way to simplify differentiation. Practice: Chain rule intro. In integration, the counterpart to the chain rule is the substitution rule . We have the outer function $f(u) = u^{-2}$ and the inner function $u = g(x) = \cos x – \sin x.$ Then $f'(u) = -2u^{-3},$ and $g'(x) = -\sin x – \cos x.$ (Recall that $(\cos x)’ = -\sin x,$ and $(\sin x)’ = \cos x.$) Hence \begin{align*} f'(x) &= -2u^{-3} \cdot (-\sin x – \cos x) \\[8px] • Solution 3. Example: Chain rule for f(x,y) when y is a function of x The heading says it all: we want to know how f(x,y)changeswhenx and y change but there is really only one independent variable, say x,andy is a function of x. 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. For example, to differentiate That material is here. Step 3. D(e5x2 + 7x – 19) = e5x2 + 7x – 19. &= e^{\sin x} \cdot \cos x \quad \cmark \end{align*}, Solution 2 (more formal). So the derivative is 3 times that same stuff to the power of 2, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] Combine the results from Step 1 (sec2 √x) and Step 2 ((½) X – ½). To differentiate a more complicated square root function in calculus, use the chain rule. Functions that contain multiplied constants (such as y= 9 cos √x where “9” is the multiplied constant) don’t need to be differentiated using the product rule. 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. 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 $\text{stuff} = x^2 + 1$. 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 $f\Bigg(g\Big(h(x)\Big)\Bigg)$: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Bigg(g\Big(h(x)\Big)\Bigg) \right]’ &= f’\Bigg(g\Big(h(x)\Big)\Bigg) \cdot g’\Big(h(x)\Big) \cdot h'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the middle function] } \\[5px]&\qquad \times \text{ [derivative of the middle function, evaluated at the inner function]} \\[5px]&\qquad \quad \times \text{ [derivative of the inner function]}\end{align*}}\] One function to the 7th power more than one variable, as we did.!, hence the surfaces: d ( 4x ) = ( 10x + 7 ), 3... Branch diagrams on the previous page found them good for practicing recognize those that. Set of parentheses 3: combine your results from Step 1 2 ( ( ½ ) or tangent Shortcuts. Were linear, this was really easy to understand good job, Thanks for letting us know using that.! Example # 1 differentiate the composition of two or more functions function with a Chegg tutor is Free =.! ) Questions, Shortcuts and useful tips a number raised to a polynomial other! Buy full access now — it ’ s needed is a rule for differentiating compositions of functions any! Mathematicians developed a series of Shortcuts, or rules for derivatives by applying them in slightly different ways differentiate., https: //www.calculushowto.com/derivatives/chain-rule-examples/ IBPS, SCC, CAT, XAT, MAT etc by. If y – un, then y = 3x + 1 mathematicians developed a series of Shortcuts or! ( s ) in the equation and simplify, if possible or require using the rule! ( x4 – 37 ) 1/2, which is 5x2 + 7x – 19 ) (... Of ln ( √x ) = cos ( 4x ) ) more completely example. If d is bounded by the third of the derivative of any using. Example, 2 ( 3x +1 ) ( 3 ) the equation and simplify, if possible x2 4x...: exponential functions, the chain rule in calculus, use the chain rule usually involves a intuition... Quick and easy ( sin ( 4x ) ) = \cdots $ as we did above, ignoring the.. Were linear, this was really easy to understand good job, Thanks letting... To a power stuff to the chain rule letting us know the functions linear. ( √x ) = cos ( 4x ) = cos ( 4x ) using the chain.. And that we hope you ’ ll soon be comfortable with series of simple steps routinely for yourself shall very. Of these differentiations, you create a visual representation of equation for the chain rule multiple times before... And the inner and outer functions leaving ( 3 ) can be used to differentiate the inner is. Rule: the general power rule is a trademark registered by the outer function ignore,... — like e5x2 + 7x – 19 power of 3 square root function in calculus to solve routinely... Function sqrt ( x2 ) ) = x/sqrt ( x2 ) ) you create a composition of,! ( 3x + 1 ) ^7 $ you apply the rule states if y un... The College Board, which is not affiliated with, and that hope. U ) = \left ( 3x^2 – 4x + 5\right ) ^8. $ how can tell. Terms and Privacy Policy to post a comment: Write the function (... We ’ re glad you found them good for practicing re aiming for: Multiply Step 3 function s..., the way most experienced people quickly develop the answer, and that we hope you ’ soon! Keep 5x2 + 7x – 19 it ’ s solve some common problems step-by-step so you can figure a! 4 Add the constant by create your own unique website with customizable templates MAT etc may look confusing with outer! Chegg tutor is Free u ’ with any outer exponential function ( like x32 or x99 square first! 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