If $f (x) = {tan}^{24} x$ $f(x) =tan^24x$ and $g (x) = \sqrt{5 x - 1}$ $g(x) = sqrt(5x-1$ , what is $f'(g(x))$ ?

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Answer 1 · 2016-07-03T06:00:16

$f'(g(x))=24tan^23(sqrt(5x-1)) xx sec^2(sqrt(5x-1))$

As $f(x)=tan^24x$ and $g(x)=sqrt(5x-1)$

As $f(x)=tan^24x$

$(df)/(dx)=f'(x)=24tan^23x xx sec^2x$

and $f'(g(x))=24tan^23(sqrt(5x-1)) xx sec^2(sqrt(5x-1))$

Answer 2 · 2016-07-04T06:03:31

Reqd. Deri. $=(48/5)tan^23x*sec^2x*sqrt(5x-1).$

$f(x)=tan^24x, g(x)=sqrt(5x-1)=t,$ say.

Now Reqd. Deri. $=f'(g(x))=f'(t)=d/dt{f(t)}=(df)/dt$

We see that $f$ is a fun. of $x$ , and $x$ of $t$ . Hence,

reqd. Deri. $=(df)/dt=(df)/dx*dx/dt..........(1)$

Now, $f(x)=tan^24x rArr (df)/dx=24tan^23x*d/dxtanx=24tan^23x*sec^2x$ .....(2)

Next, $t=sqrt(5x-1) rArr dt/dx=1/(2sqrt(5x-1))*d/dx(5x-1)=5/(2sqrt(5x-1)$ $rArr dx/dt=(2sqrt(5x-1))/5....................(3)$

From $(1),(2),(3)$ , Reqd. Deri. $=24tan^23x*sec^2x*(2sqrt(5x-1))/5$ ,
$=(48/5)tan^23x*sec^2x*sqrt(5x-1).$

If f(x) =tan^24x f(x)=tan24xf(x) =tan^24x and g(x) = sqrt(5x-1 g(x)=√5x−1g(x) = sqrt(5x-1 , what is f'(g(x)) f'(g(x)) ?