Calculus problem #2518

Let $g(x) = x+1$ and $h(x)=x ^ \frac{3}{2}$

So, by the chain rule, we have:

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

$g'(x)=1$

$h'(x)=\frac{3\sqrt{x}}{2}$

$f'(x)=\frac{3\sqrt{x+1}}{2}\times 1$

$f'(15)=\frac{3\sqrt{15+1}}{2}=\frac{3\sqrt{16}}{2}=\frac{3\times4}{2}=3\times 2= \boxed{6}$

6

f'(x)=(3 sqrt(x+1))/2---> f'(15)=(3 sqrt(16))/2=3(4)/2=6

Derivitive f(x) first and sub f'(15)

solution is f '(x) = 3/2 . (x+1)^(1/2) >> f ' (15) = 3/2 . 16^(1/2 ) = 3/2 . 4 = 6. Answer : 6 . Happy Christmas and New Year ^_^

the derivative of f(x) is f'(x) = 3/2(x+1)^(1/2) substitute x F(15) = 3/2(15+1)^(1/2) f(15) = 3/2 (16)^(1/2) f(15) = 3/2 (4) f(15) = 12/2 f(15) = 6