Chain Rule

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

$2u^1-0$   Power & Constant

$2u$

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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

Step 1: Simplify

(5x + 3)² = (5x + 3)(5x + 3)

25x² + 15x + 15x + 9

25x² + 30x + 9

Step 2: Differentiate without the chain rule.

$\frac{d}{dx}\left(25x^2+30x+9\right)$ Original

$\frac{d}{dx}25x^2+\frac{d}{dx}30x+\frac{d}{dx}9$ 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

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

$-\frac{1}{2}{u}^{-\frac{3}{2}}$

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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.

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

$-\frac{1}{2}{u}^{-\frac{3}{2}} ·2x-1$ Chain Rule

$\left[-\frac{1}{2}{\left(x^2-x+14\right)}^{-\frac{3}{2}}\right]·\left(2x-1\right)$ Sub

$\left[-\frac{1}{2}{\left(x^2-x+14\right)}^{-\frac{3}{2}}\right]·\left(2x-1\right)$

Step 1: Identify the inner function and rewrite the outer function replacing the inner function by the variable u.

$g=\left(\frac{1-2x^3}{x}\right)$ Inner Function

$u=\left(\frac{1-2x^3}{x}\right)$ 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

$2u$

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Derivative of $g=\frac{1-2x^3}{x}$

$\frac{d}{dx}\left(\frac{1-2x^3}{x}\right)$ Original

Apply Chain Rule:

$\frac{\left[\left(x\right)\frac{d}{dx}\left(1-2x^3\right)\right]-\left[\left(1-2x^3\right)\frac{d}{dx}\left(x\right)\right]}{x^2}$

$\frac{\left[\left(x\right)\left(-6x^2\right)\right]-\left[\left(1-2x^3\right)\left(1\right)\right]}{x^2}$  Take deriv.

$\frac{\left[-6x^3\right]-\left[1-2x^3\right]}{x^2}$   Simplify

$\frac{-4x^3-1}{x^2}$

Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify.

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

$2u · \frac{-4x^3-1}{x^2}$   Chain Rule

$2\left(\frac{1-2x^3}{x}\right) · \frac{-4x^3-1}{x^2}$ Sub for u

$\frac{\left(2-4x^3\right)\left(-4x^3-1\right)}{x^3}$