calc 1 question

**weezie23** · June 2nd 2008, 05:32 PM

question is f(3) =2, f'(3)=4 g(3)=8 g'(3)=3 Find h'(3)=fx/gx

does this relate to dy/dx=dy/du*du/dx....?

**topsquark** · June 2nd 2008, 05:41 PM

Originally Posted by weezie23

question is f(3) =2, f'(3)=4 g(3)=8 g'(3)=3 Find h'(3)=fx/gx

does this relate to dy/dx=dy/du*du/dx....?

If I'm reading your problem correctly that is exactly it. So h'(3) = 4/3.

-Dan

**Soroban** · June 2nd 2008, 06:24 PM

Hello, weezie23!

I'm guessing that you left out part of the problem.

$f(3) \:=\:2,\quad f'(3)\:=\:4,\quad g(3)\:=\:8,\quad g'(3)\:=\:3,\quad h(x) \:=\:\frac{f(x)}{g(x)}$

Find $h'(3)$

We have: . $h(x) \:=\:\frac{f(x)}{g(x)}$

. . Quotient Rule: . $h'(x) \;=\;\frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$

Since $h'(3) \;=\;\frac{g(3)f'(3) - f(3)g'(3)}{[g(3)]^2}$

. . we have: . $h'(3) \;=\;\frac{8\!\cdot\!4 - 2\!\cdot\!3}{8^2} \;=\;\frac{26}{64}\;=\;\frac{13}{32}$

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