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

2 Answers
Jul 3, 2016

Answer:

#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

Answer:

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).#