How do you differentiate $f (x) = csc (ln (\frac{1}{x^{4}}))$ $f(x)=csc(ln(1/x^4))$ using the chain rule?

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How do you differentiate $f (x) = csc (ln (\frac{1}{x^{4}}))$ $f(x)=csc(ln(1/x^4))$ using the chain rule?

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Answer 1 · 2017-05-21T03:43:15

$\frac{4 csc (4 ln x) cot (4 ln x)}{x}$ $(4csc(4lnx)cot(4lnx))/x$

Explanation:

This can be rewritten as:
$- csc (4 ln x)$ $-csc(4lnx)$

The derivative of:
$csc (u) = - csc (u) cot (u)$ $csc(u)=-csc(u)cot(u)$
$ln u = \frac{1}{u}$ $ln u= 1/u$

Solving by the chain rule:
$\frac{d}{d x} (- csc (4 ln x)) = csc (4 ln x) cot (4 ln x) (\frac{d}{d x} [4 ln x])$ $d/dx(-csc(4lnx))=csc(4lnx)cot(4lnx)(d/dx[4lnx])$
$= csc (4 ln x) cot (4 ln x) (\frac{4}{x})$ $=csc(4lnx)cot(4lnx)(4/x)$
$= \frac{4 csc (4 ln x) cot (4 ln x)}{x}$ $=(4csc(4lnx)cot(4lnx))/x$

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How do you differentiate $f (x) = csc (ln (\frac{1}{x^{4}}))$ $f(x)=csc(ln(1/x^4))$ using the chain rule?

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How do you differentiate $f (x) = csc (ln (\frac{1}{x^{4}}))$ $f(x)=csc(ln(1/x^4))$ using the chain rule?