Thread: Suppose that f and g are differentiable functions with the values given below...?

1. Suppose that f and g are differentiable functions with the values given below...?

Suppose that f and g are differentiable functions with the values given below. If h(x)= f {g[sqrt(x)]} and k(x)= g[ 1/x ・f(2x)], show that h'(4)=3 and k'(1)=16..

x=2, f(x)=2, g(x)=2, f'(x)=3, g'(x)=4

I have no idea with this question.. Can anyone help me?? Please!!!

2. Re: Suppose that f and g are differentiable functions with the values given below...?

what is your question and i don't understand clearly on k(x)= g[ 1/xf(2x)],

3. Re: Suppose that f and g are differentiable functions with the values given below...?

Originally Posted by Yukina
Suppose that f and g are differentiable functions with the values given below. If h(x)= f {g[sqrt(x)]} and k(x)= g[ 1/xf(2x)], show that h'(4)=3 and k'(1)=16..

x=2, f(x)=2, g(x)=2, f'(x)=3, g'(x)=4

I have no idea with this question
heard of the chain rule?

$h(x) = f[g(\sqrt{x})]$

$h'(x) = f'[g(\sqrt{x})] \cdot g'(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$

evaluate $h'(4)$

please clarify the expression for $k(x)$ ... what you posted is a mess.

4. Re: Suppose that f and g are differentiable functions with the values given below...?

Hello, Yukina!

Suppose that $f(x)$ and $g(x)$ are differentiable functions

with these values: . $\begin{Bmatrix}f(2) &=& 2 \\ g(2) &=& 2 \\ f'(2) &=&3 \\ g'(2) &=& 4 \end{Bmatrix}$

$\text{If }h(x)\:=\:f[g(\sqrt{x})]\,\text{ and }\,k(x) \:=\: g\left[\tfrac{1}{x}\!\cdot\!f(2x)],$

. . $\text{ show that: }\: h'(4)\:=\:3\,\text{ and }\,k'(1)\:=\:16$

$\text{Find }h'(x)\!:\;h'(x) \;=\;f'[g(\sqrt{x})]\cdot g'(\sqrt{x}) \cdot \tfrac{1}{2}x^{-\frac{1}{2}}$

$\text{Then: }\:h'(4) \;=\;f'[g(\sqrt{4})]\cdot g'(\sqrt{4})\cdot \frac{1}{2}\cdot\frac{1}{\sqrt{4}}$

. . . . . . . . . $=\;f'(g(2)] \cdot g'(2) \cdot\frac{1}{2}\cdot\frac{1}{2}$

. . . . . . . . . $=\;f'(2) \cdot 4\cdot \frac{1}{4}$

. . . . . . . . . $=\;3\cdot 4\cdot \frac{1}{4}$

. . . . . . . . . $=\;3$

$\text{Find }k'(x)\!:\;k'(x) \;=\;g'\left[\tfrac{1}{x}\!\cdot\!f(2x)\right]\!\cdot\!\left[\tfrac{1}{x}\!\cdot\!f'(2x)\!\cdot\!2 - \tfrac{1}{x^2}\!\cdot\!f(2x)\right]$

$\text{Then: }\:k'(1) \;=\;g'[\tfrac{1}{1}\!\cdot\!f(2)]\!\cdot\!\left[\tfrac{1}{1}\!\cdot\!2\!\cdot\! f'(2) - \tfrac{1}{1}\!\cdot\!f(2)\right]$

. . . . . . . . . $=\;g'(2)\left[2\!\cdot\!f'(2) - f(2)\right]$

. . . . . . . . . $=\;4\!\cdot\!\left[2\!\cdot\!3 - 2\right]$

. . . . . . . . . $=\;4\!\cdot\!4$

. . . . . . . . . $=\;16$