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Q. Let f and g be differentiable functions such that $f(3) = 5, g(3) = 7, f’(3) = 13, g’(3) = 6, f'(7) = 2$ and $g'(7) = 0$. If $h(x) = (fog) (x)$, then $h'(3) =$

KEAMKEAM 2018

Solution:

$h(x)=f(g(x))$
$h'(x) =f'(g(x)) \cdot g'(x)$
$h'(3) =f'(g(3)) \cdot g'(3)$
$\left[\because g(3)=7\,g'(3)=6\right]$
$ =f'(7) \cdot 6$
$=2 \times 6=12$