If $f (x) = {tan}^{2} (\frac{x}{2})$ $f(x) =tan^2(x/2)$ and $g (x) = \sqrt{5 x - 1}$ $g(x) = sqrt(5x-1$ , what is $f'(g(x))$ ?

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Answer 1 · 2016-03-16T13:05:20

$f'(g(x))=tan(sqrt(5x-1)/2)xxsec^2(sqrt(5x-1)/2)$

As $f(x)=tan^2(x/2)$ ,

$f'(x)=(d(tan^2(x/2)))/(d(tanx/2))$ $xx$ $(d(tanx/2))/(d(x/2))$ $xx$ $(d(x/2))/dx$

Hence $f'(x)=2tan(x/2)xxsec^2(x/2)xx1/2$

or $f'(x)=tan(x/2)xxsec^2(x/2)$

Hence $f'(g(x))=tan(g(x)/2)xxsec^2(g(x)/2)$

and as $g(x)=sqrt(5x-1)$

$f'(g(x))=tan(sqrt(5x-1)/2)xxsec^2(sqrt(5x-1)/2)$

If f(x) =tan^2(x/2) f(x)=tan2(x2)f(x) =tan^2(x/2) and g(x) = sqrt(5x-1 g(x)=√5x−1g(x) = sqrt(5x-1 , what is f'(g(x)) f'(g(x)) ?