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the Chain Rule (for Derivatives) |
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| Typical Application Example: | Calculate the derivative of this function | \( \sin(3x^{1/5}-2x^3) \) | ||||
| Chain Rule | \[ \frac{d}{dx}\,\Big( g \big(h(x)\big)\Big) = g'\big(\,h(x)\,\big)\,\cdot\, h'(x) \] | |||||
| the Reason for its existence: | It computes the derivative of composed functions, \( g \big(\, h( x )\, \big) \) | |||||
| Using the Chain Rule | First, identify the composition, \( g \) and \( h(x) \) | Second, calculate \( g'\big(\,h(x)\,\big)\,\cdot\, h'(x) \) | ||||
| First Identify |
the composition parts: the inside and the outside of \[ g\,\big(\, h(x) \,\big)\, \] |
the outside function is \( g \), and the inside function is \( h(x) \). |
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| outside | \[ g\,\big(\, h(x) \,\big)\, = \sin \big( 3x^{1/5}-2x^3 \big) \] | \( g(x)=\) | \( \sin(x) \) | |||
| inside | \[ g\,\big(\, h(x) \,\big)\, = \sin \big( 3x^{1/5}-2x^3 \big) \] | \( h(x)= \) | \( 3x^{1/5}-2x^3 \) | |||
| Calculate using Chain Rule Formula |
formula or pattern |
\[g'\,\big(\,h(x)\,\big)\,\cdot\, h'(x) \] |
the chain rule pattern uses \( g' \) and \( h' \) so we need to calculate them, |
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| the derivative | \( g'(x)=\) | \( \cos(x) \) | and | \( h'(x)= \) | \( 3\cdot \frac{1}{5} x^{-4/5}-2\cdot 3 x^2 \) | |
| Finally, use the pattern. | First: | \( g'\big( h(x) \big) = \cos\big( h(x) \big) = \cos\big( 3x^{1/5}-2x^3 \big) \) | ||||
| then multiply by \( h'(x) \) | \( g' \big(h(x)\big)\cdot h'(x) = \cos\big(3x^{1/5}-2x^3\big)\,\cdot\, \big( \frac{3}{5}x^{-4/5}-6x^2 \big) \). | |||||
| The Answer: | \[ \frac{d}{dx}\big( \sin(3x^{1/5}-2x^3) \big) = \cos\big(3x^{1/5}-2x^3\big)\,\cdot\, \big( \frac{3}{5}x^{-4/5}-6x^2 \big) \] | |||||
| \[ \frac{d}{dx}\big( \sin(3\sqrt[5]{x}-2x^3) \big) = \cos\big(3\sqrt[5]{x}-2x^3\big)\,\cdot\, \Big( \frac{3}{5 \sqrt[5]{x^4}}-6x^2 \Big) \] | reduce to taste | |||||