Question

Suppose that g(3)=1, g'(3)=2 and f'(-1)= 0.5 find f(g(x))'3

Answer 1

I think the question is wrong. It will be either $g(3)=-1$ or $f'(1)=0.5.$ Let $F(x)=f(g(x))$ .

Then $F'(x)=\frac {d}{dx}[f(g(x))]$

$=f'(g(x)).\frac{d}{dx}[g(x)]$

$=f'(g(x)).g'(x)$

Therefore, $F'(3)=f'(g(x))(3)$

$=f'(g(3)).g'(3)$

$=2.f'(-1)$ [ taking $g(3)=-1$ ]

$=2.\frac{1}{2}$ [As $f'(-1)=0.5$ ]

$=1$

$\therefore f(g(x))'(3)=1.$