Calculus - Composite Function Derivation

Enterprise 2023-08-12 21:23:01 views: null

Learn now: https://edu.csdn.net/course/play/26262/326627?utm_source=blogtoedu

Derivative of function multiplication:

prove:(\mu\upsilon)'=\mu'\upsilon+\upsilon'\mu

P：

\begin{aligned} (\mu\upsilon)' &=\lim_{\triangle x \to 0} \frac{(\mu\upsilon)(x+\triangle x)-(\mu\upsilon)(x)}{\triangle x} \\ &=\lim_{\triangle x \to 0} \frac{\mu(x+\triangle x)\upsilon(x+\triangle x)-\mu(x)\upsilon(x)}{\triangle x}\\ &=\lim_{\triangle x \to 0} \frac{\mu(x+\triangle x)\upsilon(x+\triangle x)-\mu(x)\upsilon(x) +\mu(x+\triangle x)\upsilon(x)-\mu(x+\triangle x)\upsilon(x)}{\triangle x}\\ &=\lim_{\triangle x \to 0} \frac{\mu(x+\triangle x)-\mu(x)}{\triangle x}\upsilon(x) +\lim_{\triangle x \to 0} \frac{\upsilon(x+\triangle x)-\upsilon(x)}{\triangle x}\mu(x+\triangle x)\\ &=\mu'\upsilon+\upsilon'\mu \end{aligned}

 

 

Derivative of function division:

Q：(\frac{\mu}{\upsilon})'=?

A：

\begin{aligned} \triangle (\frac{\mu}{\upsilon}) &=\frac{\mu (x+\triangle x)}{\upsilon (x+\triangle x)} -\frac{\mu (x)}{\upsilon (x)}\\ &=\frac{\upsilon \triangle \mu-\mu \triangle \upsilon}{\upsilon(\upsilon+\triangle \upsilon)} \end{aligned}

\begin{aligned} (\frac{\mu}{\upsilon})' &=\lim_{\triangle x \to 0} \frac{\triangle \frac{\mu}{\upsilon}}{\triangle x}\\ &=\lim_{\triangle x \to 0} \frac{\upsilon \frac{\triangle \mu}{\triangle x}-\mu\frac{\triangle \upsilon }{\triangle x}} {\upsilon(\upsilon+\triangle \upsilon)}\\ &=\frac{\mu'\upsilon-\mu\upsilon'}{\upsilon^2} \end{aligned}

 

Chain rule:

\frac{\mathrm{d}}{\mathrm{d} t} f(g(t))=f'(g(t))g'(t)

 

(LaTeX formula online editor https://www.latexlive.com/ )

 

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