Differentiate $\sin(2x)/x^2$? this one got some scratching my head.
Having trouble getting the dy/dx for the equation.
Anyone has an answer.
An explanation would be appreciated so I know the steps to solve similar equations.
 A: Hint by way of quotient rule: If $f(x)$ and $g(x)$ are differentiable, $g(x)\ne 0$, then 
$$
\left(\frac{f(x)}{g(x)} \right)'=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}
$$
for you, $f(x)=\sin(2x)$ and $g(x)=x^2$.
A: $f(x) = \frac{sin(2x)}{x^2}$, using the quotient rule to obtain:
$$\frac{dy}{dx} = \frac{2x^2\cos(2x)-2x\sin(2x)}{x^4}$$$$\frac{dy}{dx} = \frac{2x\cos(2x)-2\sin(2x)}{x^3}$$
A: As an alternative by product rule $fg=f'g+fg'$ with


*

*$f(x)=\sin 2x \implies f'(x)=2\cos 2x$

*$g(x)=\frac{1}{x^2}\implies g'(x)=\frac{-2}{x^3}$

A: Just another way to do it.
When you have expressions which just involve products, quotients and powers, logarithmic differentiation makes life simpler
$$y=\frac{\sin(ax)}{x^b}\implies \log(y)=\log(\sin(ax))-b\log(x)$$ Differentiate both sides
$$\frac{y'}y=\frac{a \cos(ax)}{\sin(ax)}-\frac b x$$ Now
$$y'=y \times \left(\frac{y'}y\right)$$ and simplify as much as you can.
A: Simply apply chain rule and  the rule for $(\frac fg)'=\frac {f'g-fg'}{g^2}$
Note that $(\sin(2x))'=2\cos(2x)$ and $(x^2)'=2x$
$$y=\frac {\sin(2x)}{x^2}$$
$$y'=\frac { 2x^2\cos(2x)-\sin(2x)2x}{x^4}$$
Then simplify the expression
$$....$$
A: Use the rule
$$\left(\dfrac{f}{g}\right)' = \dfrac{f'g-fg'}{g^2}$$
Substituting $f(x)=\sin(2x)$ and $g(x)=x^2$, we have
$$\left(\dfrac{\sin(2x)}{x^2}\right)' = \dfrac{2\cos(2x)x^2-\sin(2x)2x}{x^4}=
\dfrac{2x\cos(2x)-2\sin(2x)}{x^3}
$$
