# Chain Rule

This discussion will focus on the Chain Rule of Differentiation. The chain rule allows the differentiation of composite functions, notated by $f\circ g$. For example take the composite function (x + 3)2. The inner function is g = x + 3. If x + 3 = u then the outer function becomes f = u2.

This rule states that:

The derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function

CHAIN RULE OF DIFFERENTIATION:

$\left(f\circ g\right)\left(x\right)={f}^{\prime }\left(g\left(x\right)\right)·g\prime \left(x\right)$

Let's work some examples

To work these examples requires the use of various differentiation rules. If you are not familiar with a rule go to the associated topic for a review.

(5x + 3)2
 Step 1: Identify the inner function and rewrite the outer function replacing the inner function by the variable u. g = 5x + 3 Inner Function u = 5x + 3 Set Inner Function to u f = u2        Outer Function Step 2: Take the derivative of both functions. Derivative of f = u2 $\frac{d}{dx}{u}^{2}$      Original $2{u}^{1}-0$   Power & Constant $2u$ ______________________________ Derivative of g = x + 3 $\frac{d}{dx}5x+3$ Original $\frac{d}{dx}5x+\frac{d}{dx}3$ Sum Rule $5\frac{d}{dx}x+3$  Constant Multiple 5x0 + 0      Power & Constant 5 Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify. $\left(f\circ g\right)\left(x\right)={f}^{\prime }\left(g\left(x\right)\right)·g\prime \left(x\right)$ 2u(5)       Chain Rule 2(5x + 3)(5) Substitute for u 50x + 30   Simplify If the expression is simplified first, the chain rule is not needed. Step 1: Simplify (5x + 3)2 = (5x + 3)(5x + 3) 25x2 + 15x + 15x + 9 25x2 + 30x + 9 Step 2: Differentiate without the chain rule. $\frac{d}{dx}\left(25{x}^{2}+30x+9\right)$ Original $\frac{d}{dx}25{x}^{2}+\frac{d}{dx}30x+\frac{d}{dx}9$ Sum Rule $25\frac{d}{dx}{x}^{2}+30\frac{d}{dx}x+\frac{d}{dx}9$ Constant Multiple 25(2x1) + 30x0 + 0    Power & Constant 50x + 30
Example 1:      $\frac{1}{\sqrt{{x}^{2}-x+14}}$
 Step 1: Identify the inner function and rewrite the outer function replacing the inner function by the variable u. g = x2 - x + 14 Inner Function u = x2 - x + 14 Set IF to u $f={{u}}^{-\frac{1}{2}}$     Outer Function Step 2: Take the derivative of both functions. Derivative of $f={u}^{-\frac{1}{2}}$ $\frac{d}{dx}{u}^{-\frac{1}{2}}$ Original $-\frac{1}{2}{u}^{-\frac{3}{2}}$ Power $-\frac{1}{2}{u}^{-\frac{3}{2}}$ __________________________ Derivative of g = x2 - x + 14 $\frac{d}{dx}\left({x}^{2}-x+14\right)$ Original $\frac{d}{dx}{x}^{2}-\frac{d}{dx}x+\frac{d}{dx}14$  Sum/Diff Rule 2x1 - 1x0 + 0 Power/Constant $2x-1$ Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify. $\left(f\circ g\right)\left(x\right)={f}^{\prime }\left(g\left(x\right)\right)·g\prime \left(x\right)$ Chain Rule $\left[-\frac{1}{2}{\left({x}^{2}-x+14\right)}^{-\frac{3}{2}}\right]·\left(2x-1\right)$ Sub $\left[-\frac{1}{2}{\left({x}^{2}-x+14\right)}^{-\frac{3}{2}}\right]·\left(2x-1\right)$
Example 2:      ${\left(\frac{1-2{x}^{3}}{x}\right)}^{2}$
 Step 1: Identify the inner function and rewrite the outer function replacing the inner function by the variable u. $g=\left(\frac{1-2{x}^{3}}{x}\right)$ Inner Function $u=\left(\frac{1-2{x}^{3}}{x}\right)$ Set IF to u f = u2      Outer Function Step 2: Take the derivative of both functions. Derivative of f = u2 $\frac{d}{dx}{u}^{2}$     Original 2u1        Power $2u$ ________________________ Derivative of ${g}{=}\frac{1-2{x}^{3}}{x}$ $\frac{d}{dx}\left(\frac{1-2{x}^{3}}{x}\right)$ Original Apply Chain Rule: $\frac{\left[\left(x\right)\frac{d}{dx}\left(1-2{x}^{3}\right)\right]-\left[\left(1-2{x}^{3}\right)\frac{d}{dx}\left(x\right)\right]}{{x}^{2}}$ $\frac{\left[\left(x\right){\left(}-6{x}^{2}{\right)}\right]-\left[\left(1-2{x}^{3}\right){\left(}{1}{\right)}\right]}{{x}^{2}}$  Take deriv. $\frac{\left[-6{x}^{3}\right]-\left[1-2{x}^{3}\right]}{{x}^{2}}$   Simplify $\frac{-4{x}^{3}-1}{{x}^{2}}$ Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify. $\left(f\circ g\right)\left(x\right)={f}^{\prime }\left(g\left(x\right)\right)·g\prime \left(x\right)$ Chain Rule Sub for u $\frac{\left(2-4{x}^{3}\right)\left(-4{x}^{3}-1\right)}{{x}^{3}}$

 Related Links: Math algebra Trigonometry Differentiation Rules Inverse Trigonometric Differentiation Rules Calculus Topics

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