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\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
eg:
| \(x\) | 1 | 2 | 3 |
|---|---|---|---|
| \(f(x)\) | 3 | 2 | 1 |
| \(f'(x)\) | 4 | 5 | 6 |
\[ g'(x) = \begin{cases} 2 & 0 < x < 2 \\ -4 & x > 2 \\ \text{DNE} & x = 2 \end{cases} \]
\( h(x) = f(g(x)) \)
\( k(x) = g(f(x)) \)
Find \( h'(1) \) and \( k'(3) \)
\[ h'(x) = f'(g(x)) \, g'(x) \rightsquigarrow h'(1) = f'(g(1)) \cdot g'(1) \]
\[ = f'(2) \cdot 2 = 5 \cdot 2 = 10 \]
\[ k'(x) = g'(f(x)) \cdot f'(x) \rightsquigarrow k'(3) = g'(f(3)) \cdot f'(3) = g'(1) \cdot 6 = 2 \cdot 6 = 12 \]
eg: \( y = e^{5x^2-8} \)
\[ \begin{aligned} y' = \frac{d}{dx}\left[ e^{\overbrace{5x^2-8}^{u}} \right] &= \frac{d}{dx} e^u \\ &= \frac{d}{du} e^u \cdot \frac{du}{dx} \\ &= e^u \frac{d}{dx} u \\ &= e^{5x^2-8} \cdot \frac{d}{dx}(5x^2-8) \\ &= 10x \, e^{5x^2-8} \end{aligned} \]
Given the function:
\[ y = \sin(e^{3x}) \]
Differentiating with respect to \( x \):
\[ y' = \frac{d}{dx} \sin(e^{3x}) \qquad (u = e^{3x}) \] \[ = \frac{d}{dx} \sin u \]
Applying the Chain Rule:
\[ = \frac{d \sin u}{du} \frac{du}{dx} \] \[ = \cos u \frac{du}{dx} \] \[ = \cos(e^{3x}) \underbrace{\frac{d}{dx} e^{3x}}_{\text{use Chain Rule Again}} \qquad (v = 3x) \] \[ = \cos(e^{3x}) \frac{d e^v}{dx} \frac{dv}{dx} \] \[ = \cos(e^{3x}) \, e^{3x} \cdot 3 \]
Reference Derivative Formulas:
Problem:
Differentiate \( y = \operatorname{cosec}[\cos(x^3+5)] \)
Solution:
\[ y' = \frac{d}{dx} \operatorname{cosec}[\cos(x^3+5)] \]
Let \( u = \cos(x^3+5) \):
\[ = \frac{d}{dx} \operatorname{cosec} u = \frac{d}{du} \operatorname{cosec} u \cdot \frac{du}{dx} \] \[ = -\operatorname{cosec} u \cot u \cdot \frac{du}{dx} \] \[ = -\operatorname{cosec}(\cos(x^3+5)) \cot(\cos(x^3+5)) \cdot \frac{d}{dx} \cos(\underbrace{x^3+5}_{v}) \]
Apply chain rule again with \( v = x^3+5 \):
\[ = -\operatorname{cosec}(\cos(x^3+5)) \cot(\cos(x^3+5)) \cdot \frac{d}{dv}\cos v \cdot \frac{dv}{dx} \] \[ = \left(-\operatorname{cosec}(\cos(x^3+5)) \cot(\cos(x^3+5))\right) \left(-\sin(x^3+5)\right) \cdot 3x^2 \]
Chain Rule: \( \text{inside} = u = \sec(x^2) \cdot (x^3 + 9) \)
\[ y' = \frac{d}{dx} u^{3/4} = \frac{3}{4} u^{-1/4} \cdot \frac{du}{dx} \]
Using the Product Rule to find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{d}{dx} \left[ \sec(x^2) \cdot (x^3 + 9) \right] = \frac{d}{dx}\sec(x^2) \cdot (x^3 + 9) + \sec(x^2) \cdot \underbrace{\frac{d}{dx}(x^3 + 9)}_{3x^2} \]
Use Chain Rule for \( \frac{d}{dx}\sec(x^2) \):
\( = \text{outside}' \cdot \text{inside}' = \sec(x^2) \tan(x^2) \cdot \frac{d}{dx} x^2 = 2x \sec(x^2) \tan(x^2) \)
\[ \frac{du}{dx} = 2x \sec(x^2) \tan(x^2) (x^3 + 9) + 3x^2 \sec(x^2) \] \[ y' = \frac{3}{4} \left[ \sec(x^2) \cdot (x^3 + 9) \right]^{-1/4} \left( 2x \sec(x^2) \tan(x^2) (x^3 + 9) + 3x^2 \sec(x^2) \right) \]
\[ y = \frac{\tan(x^3+5)}{e^{\sin x}} \]
\[ N = \tan(x^3+5), \quad D = e^{\sin x} \]
\[ \frac{d}{dx} \left[\frac{N}{D}\right] = \frac{N'D - ND'}{D^2} \]
\[ N' = \frac{d}{dx} \tan(x^3+5) = \frac{d}{du} \tan u \cdot \frac{du}{dx}, \quad u = x^3+5 \]
\[ = (\sec^2 u) \cdot 3x^2 = 3x^2 \sec^2(x^3+5) \]
\[ D' = \frac{d}{dx} e^{\sin x} = \frac{d}{dv} e^v \cdot \frac{dv}{dx}, \quad v = \sin x \]
\[ = e^{\sin x} \cdot \cos x \]
\[ y' = \frac{3x^2 \sec^2(x^3+5) e^{\sin x} - \tan(x^3+5) e^{\sin x} \cos x}{(e^{\sin x})^2} \]
Example: \( g(x) = \cos(\pi f^2(x)), \quad f(1) = \frac{1}{2}, \quad f'(1) = 3 \quad \text{Find } g'(1) \)
\[ \text{inside} = \pi [f(x)]^2 \] \[ (\text{inside})' = \pi \cdot \frac{d}{dx}[f(x)]^2 \overset{\text{use Chain Rule}}{=} \pi \frac{d}{du} u^2 \cdot \frac{du}{dx}, \quad u = f(x) \] \[ = 2\pi u \cdot \frac{du}{dx} \] \[ = 2\pi f(x) f'(x) \] \[ g'(x) = \frac{d}{dx}\cos(\pi f^2(x)) = -\sin(\pi f^2(x)) \cdot 2\pi f(x) f'(x) \] \[ g'(1) = -\sin(\pi f^2(1)) \cdot 2\pi f(1) f'(1) = -\sin\left(\frac{\pi}{4}\right) \cdot 2\pi \cdot \frac{1}{2} \cdot 3 \] \[ = \frac{-3\pi \sqrt{2}}{2} \]
\( f(x) = e^u \quad \) where \( u = e^{e^{e^x}} \)
\[ f'(x) = e^u \cdot \frac{du}{dx} = e^u \cdot \frac{d}{dx} e^{e^{e^x}} \quad \text{where } v = e^{e^x} \] \[ = e^u \cdot \frac{d}{dx} e^v \] \[ = e^u \cdot e^v \cdot \frac{d}{dx} e^{e^x} \quad \text{where } w = e^x \] \[ = e^u \cdot e^v \cdot \frac{d}{dx} e^w \] \[ = e^u \cdot e^v \cdot e^w \cdot \frac{d}{dx} e^x = e^u \cdot e^v \cdot e^w \cdot e^x \] \[ f'(1) = e^{e^{e^e}} \cdot e^{e^e} \cdot e^e \cdot e \]
eg:
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| \(f(x)\) | 4 | 0 | 7 | -8 |
| \(f'(x)\) | 2 | 1 | 2 | 8 |
| \(g(x)\) | 3 | 5 | 2 | 4 |
| \(g'(x)\) | -4 | 1 | 3 | 3 |
\(h(x) = f(g(x)), \quad p(x) = g(f(x)), \quad \text{find } h'(1) \text{ and } p'(1)\)
\[ h'(1) = f'(g(1)) \cdot g'(1) = f'(3) \cdot (-4) = 2(-4) = -8 \] \[ p'(1) = g'(f(1)) \cdot f'(1) = g'(4) \cdot 2 = 3(2) = 6. \]