Could someone explain to me step by step of how to solve the following question using the quotient rule, much appreciated (Hi)

y= 2SIN5x/x^{3}

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- Jan 27th 2014, 08:41 AMEpicAsianDudequotient rule
Could someone explain to me step by step of how to solve the following question using the quotient rule, much appreciated (Hi)

y= 2SIN5x/x^{3} - Jan 27th 2014, 09:41 AMebainesRe: quotient rule
You must memorize this equation: given $\displaystyle y(x) = \frac {f(x)}{g(x)} $ then $\displaystyle y'(x) = \frac {f'(x)g(x) -f(x)g'(x)}{(g(x))^2} $.

For your problem you have $\displaystyle f(x) = 2 \sin(5x)$ and $\displaystyle g(x) = x^3$. so:

$\displaystyle y'(x) = \frac {f'(x)g(x) -f(x)g'(x)}{(g(x))^2} = \frac {2 \times 5 \cos(5x) x^3 - 2 \sin(5x) 3 x^2}{(x^3)^2} = \frac {10 x \cos(5x)-6 \sin(5x)}{x^4}$. - Jan 27th 2014, 09:41 AMSorobanRe: quotient rule
Hello, EpicAsianDude!~

Quote:

Could someone explain to me step by step of how to find the derivative.

. . $\displaystyle y \:=\: \frac{2\sin(5x)}{x^3}$

We have: .$\displaystyle y \:=\:2\cdot\frac{\overbrace{\sin(5x)}^{f(x)}}{ \underbrace{x^3}_{g(x)}}$

Quotient Rule: .$\displaystyle y' \;=\;2\cdot\frac{\overbrace{x^3}^{g(x)}\cdot \overbrace{5\cos(5x)}^{f'(x)} - \overbrace{\sin(5x)}^{f(x)}\cdot \overbrace{3x^2}^{g'(x)}}{\underbrace{x^6}_{[g(x)]^2}}$

. . . . . . . . . . . $\displaystyle y' \;=\;2\cdot\frac{x^2\left[5x\cos(5x) - 3\sin(5x)\right]}{x^6} $

. . . . . . . . . . . $\displaystyle y' \;=\;2\cdot\frac{5x\cos(5x)-3\sin(5x)}{x^4}$ - Jan 27th 2014, 09:58 AMEpicAsianDudeRe: quotient rule
Thank you for replying ebaines & sorobans. So when answering this type of question the quotient rule formula has to be implemented in order to work out the derivative.

- Jan 27th 2014, 10:02 AMEpicAsianDudeRe: quotient rule
I have another question that i need help on but relating to the chain rule, in the following question:

y=5x^2e^2x - Jan 27th 2014, 10:17 AMebainesRe: quotient rule
- Jan 27th 2014, 10:27 AMEpicAsianDudeRe: quotient rule
So:

y= 5x^2e^2x

is equal to = 5(x^2 2e^2x + 2xe^2x) - Jan 27th 2014, 07:50 PMProve ItRe: quotient rule