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If #f(x)= tan8 x # and #g(x) = 2 x #, how do you differentiate #f(g(x)) # using the chain rule?

Answer 1

Emilia Aguilar

Derivative of #f(g(x))# is #32sec^2 16x#.

As #f(x)=tan8x#, #(df)/(dx)=sec^2 8x*8=16sec^2 8x#

and as #g(x)=2x# #(dg)/(dx)=2#

and #f(g(x))=tan8(2x)=tan16x#

and according to chain rule

#(df)/(dx)=(df)/(dg(x))*(dg)/(dx)#

= #16sec^2 8(g(x))*2#

= #16sec^2(8*2x)*2#

= #32sec^2 16x#

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Answer 2

Emily Ammerman

To differentiate ( f(g(x)) ) using the chain rule, you first find the derivative of the outer function with respect to its inner function, and then multiply it by the derivative of the inner function.

Given ( f(x) = \tan^8(x) ) and ( g(x) = 2x ),

Find ( f'(x) ): ( f'(x) = 8\tan^7(x)\sec^2(x) )
Find ( g'(x) ): ( g'(x) = 2 )
Substitute ( g(x) ) into ( f'(x) ): ( f'(g(x)) = 8\tan^7(2x)\sec^2(2x) )
Multiply by ( g'(x) ): ( f'(g(x)) \cdot g'(x) = 8\tan^7(2x)\sec^2(2x) \cdot 2 = 16\tan^7(2x)\sec^2(2x) )

So, ( \frac{d}{dx}[f(g(x))] = 16\tan^7(2x)\sec^2(2x) ).

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