Could someone explain to me step by step of how to solve the following question using the quotient rule, much appreciated
y= 2SIN5x/x^{3}
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:
$\displaystyle y'(x) = \frac {f'(x)g(x) -f(x)g'(x)}{(g(x))^2} = \frac {2 \times 5 \cos(5x) x^3 - 2 \sin(5x) 3 x^2}{(x^3)^2} = \frac {10 x \cos(5x)-6 \sin(5x)}{x^4}$.
Hello, EpicAsianDude!~
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} $
. . . . . . . . . . . $\displaystyle y' \;=\;2\cdot\frac{5x\cos(5x)-3\sin(5x)}{x^4}$