# j-james/math All my math notes, now in Markdown.

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# 3-6 Chain Rule

## Learning Targets

You will be able to

• Differentiate composite functions using the chain rule
• Find slopes of parametric functions ## The Chain Rule

If $f$ is differentiable at the point $u=g(x)$ and $g$ is differentiable at $x$, then the composite function $(f\circ g)(x) = f(g(x))$ is differentiable at $x$.

#### Newton Notation

$$(f\circ g)'(x) = f'(g(x)) \bullet g'(x)$$

#### Liebniz

If $y = f(u)$ and $u = g(x)$, then $$\frac{dy}{dx}=\frac{dy}{du}\bullet\frac{du}{dx}$$

#### In words

The derivative of the outside function evaluated at the inner function times the derivative of the inside function. ### Parametric Rule for dy/dx

If all three derivatives exist, and $\frac{dx}{dt} \neq 0$, $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$

## Examples

### Example 1

$$y = 5(x^3 + 7)^3$$ $$y = 5(g(x))^3$$ $$y' = 15(g(x))^2 \bullet g'(x)$$ $$y' = 15(x^3 + 7)^2 \bullet (3x^2 + 0)$$ $$y' = 45x^2(x^3 + 7)^2$$

### Example 2

Determine the derivative of $f(x) = (3x^2 + 1)^2$ $$f'(x) = 2(3x^2 + 1)\frac{d}{dx}(3x^2 + 1)$$ $$f'(x) = 2(3x^2 + 1)(6x + 0)$$ $$f'(x) = 12x(3x^2 + 1)$$

### Example 3

Determine the derivative of $g(t) = \tan(5-\sin 2t)$ $$g'(t) = \sec^2(5-\sin 2t)\frac{d}{dx}(5-\sin 2t)$$ $$g'(t) = \sec^2(5-\sin 2t)(-\cos 2t)\frac{d}{dx} 2t$$ $$g'(t) = \sec^2(5-\sin 2t)(-\cos 2t)(2)$$ $$g'(t) = 2\sec^2(5-\sin 2t)(-\cos 2t)$$

### Example 4

Find the derivative of the following. Note: $\frac{d}{dx}F(x) = F’(x) = f(x)$ $$f(\theta) = \cos^2 2\theta$$ $$F(x) = (3x - 1)^4(2x + 1)^{-1}$$

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