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

2 Answers
Jul 3, 2016

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

Explanation:

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))

Jul 4, 2016

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

Explanation:

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).