If $f(x)= csc 3 x$ and $g(x) = sqrt(2x-3$ , how do you differentiate $f(g(x))$ using the chain rule?

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Answer 1 · 2016-01-17T06:43:57

$f' (g(x))=-3/sqrt(2x-3)*csc (3sqrt(2x-3) )*cot (3sqrt(2x-3))$

Given $f(x)=csc 3x$ and $g(x) = sqrt(2x-3)$

$f(g(x))= csc 3g(x)$
$f(g(x))=csc 3sqrt(2x-3)$

The formula for derivative of $csc u:$

$d/dx(csc u) = - csc u*cot u* (du)/dx$

Let $u=3sqrt(2x-3)$

Take note:
$d/dx(3 sqrt(2x-3))=3* 1/(2sqrt(2x-3))*2$

$f '(g(x))=$
$- csc (3 sqrt(2x-3)) * cot (3 sqrt(2x-3)) * 3* 1/(2sqrt(2x-3))*2$

$f' (g(x))=-3/sqrt(2x-3)*csc (3sqrt(2x-3) )*cot (3sqrt(2x-3))$

If f(x)= csc 3 x f(x)= csc 3 x and g(x) = sqrt(2x-3 g(x) = sqrt(2x-3 , how do you differentiate f(g(x)) f(g(x)) using the chain rule?