#f(2)=f(4)=g'(2)=g'(4)=2# and #g(2)=g(4)=f'(2)=f'(4)#. If #h(x)=f(g(x))#, then #h'(2)#?

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Answer 1 · 2018-06-05T15:42:32

#2f^'(a)#

where #a# is the common value of

#g(2)=g(4)=f^'(2)=f^'(4)=a#
It seems that the problem is incomplete.

By the chain rule of differentiation

#h^'(x) = f^'(g(x))times g^'(x)#

Thus

#h^'(2) = f^'(g(2))times g^'(2)#
#qquad = 2f^'(a)#

where #a# is the common value of

#g(2)=g(4)=f^'(2)=f^'(4)=a#