Let f(x) and g(x) be differentiable functions | vedclass

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(C) Given $h(x) = f(g(f(x)))$.Using the chain rule for differentiation,we have:$h'(x) = f'(g(f(x))) \cdot g'(f(x)) \cdot f'(x)$.Now,we evaluate $h'(x)

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(C) Given $h(x) = f(g(f(x)))$.Using the chain rule for differentiation,we have:$h'(x) = f'(g(f(x))) \cdot g'(f(x)) \cdot f'(x)$.Now,we evaluate $h'(x)$ at $x = 2$:$h'(2) = f'(g(f(2))) \cdot g'(f(2)) \cdot f'(2)$.Given $f(2) = 1$,substitute this into the expression:$h'(2) = f'(g(1)) \cdot g'(1) \cdot f'(2)$.Given $g(1) = 2$,substitute this into the expression:$h'(2) = f'(2) \cdot g'(1) \cdot f'(2)$.Given $f'(2) = 4$ and $g'(1) = 4$,substitute these values:$h'(2) = 4 \cdot 4 \cdot 4 = 64$.

Let $f(x)$ and $g(x)$ be differentiable functions on $R$. If $h(x) = f(g(f(x)))$,where $f(2) = 1$,$g(1) = 2$ and $f'(2) = g'(1) = 4$,then $h'(2)$ is equal to:

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