# Math Help - Find dy/dx

1. ## Find dy/dx

Question : Find $\frac{dy}{dx}$

$y = \frac{(x+2)^3 (3x+5)^{-4} sinx}{(2x+2)^2}$

2. differentiate :

$\ln y = 3 \ln (x+2) - 4\ln(3x+5) + \ln(\sin x) - 2\ln (2x+2)$

3. hmm thats a neat way of interpreting a derivative thanks for that gives me a new perspective on computing really long derivatives.

4. ## I am stuck here

Originally Posted by dedust
differentiate :

$\ln y = 3 \ln (x+2) - 4\ln(3x+5) + \ln(\sin x) - 2\ln (2x+2)$
Taking log on both sides

$\ln y = 3 \ln (x+2) - 4\ln(3x+5) + \ln(\sin x) - 2\ln (2x+2)$

Differentiating wrt x

$\frac{1}{y} \ \frac{dy}{dx}$ = $\frac{9}{x+2} - \frac{32}{3x+5} + cot x - \frac{8}{2x+2}$

$\frac{dy}{dx}$ = $y \ \frac{9}{x+2} - \frac{32}{3x+5} + cot x - \frac{8}{2x+2}$ ..................I am stuck here ??????

5. Originally Posted by zorro
Taking log on both sides

$\ln y = 3 \ln (x+2) - 4\ln(3x+5) + \ln(\sin x) - 2\ln (2x+2)$

Differentiating wrt x

$\frac{1}{y} \ \frac{dy}{dx}$ = $\frac{9}{x+2} - \frac{32}{3x+5} + cot x - \frac{8}{2x+2}$

$\frac{dy}{dx}$ = $y \ \frac{9}{x+2} - \frac{32}{3x+5} + cot x - \frac{8}{2x+2}$ ..................I am stuck here ??????
remember that

$\frac{d}{dx} \ln f(x) = \frac{f'(x)}{f(x)}$

hence
$\frac{d}{dx} \{4\ln (3x + 5) \}= \frac{4 \times 3}{(3x + 5)} = \frac{12}{3x + 5}$

6. I am getting the following answer

$
\frac{dy}{dx} = y \frac{13}{(x-2)} - \frac{12}{(3x+5)} + cot x - \frac{4}{(2x+2)}
$
.....................Is this correct?????

7. Originally Posted by zorro
I am getting the following answer

$
\frac{dy}{dx} = y \frac{13}{(x-2)} - \frac{12}{(3x+5)} + cot x - \frac{4}{(2x+2)}
$
.....................Is this correct?????
I haven't checked all of your work but if
[tex]\frac{1}{y}\frac{dy}{dx} = y \frac{13}{(x-2)} - \frac{12}{(3x+5)} + cot x - \frac{4}{(2x+2)}[/itex]
then $
\frac{dy}{dx} = y (\frac{13}{(x-2)} - \frac{12}{(3x+5)} + cot x - \frac{4}{(2x+2)})$

(Note the parentheses.)

8. thanks mite

$\frac{dy}{dx} = y \left( \frac{3}{x+2} - \frac{12}{3x+5} + cot x - \frac{4}{2x+2} \right)$

9. Originally Posted by zorro
thanks mite

$\frac{dy}{dx} = y \left( \frac{3}{x+2} - \frac{12}{3x+5} + cot x - \frac{4}{2x+2} \right)$
don't forget to substitute back $y$
$\frac{dy}{dx} = \frac{(x+2)^3 (3x+5)^{-4} sinx}{(2x+2)^2} \left( \frac{3}{x+2} - \frac{12}{3x+5} + cot x - \frac{4}{2x+2} \right)$

10. Thanks mite