Differentiate $\sin(2x)/x^2$?

Question

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.

Answer 1

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 $g(x)=x^2$.

Answer 2

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 $$....$$

Answer 3

$f(x) = \frac{sin(2x)}{x^2}$, using the quotient rule to obtain: $$\frac{dy}{dx} = \frac{2x^2\cos(2x)-2x\sin(2x)}{x^4}$$$$\frac{dy}{dx} = \frac{2x\cos(2x)-2\sin(2x)}{x^3}$$

Answer 4

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 $$\left(\dfrac{\sin(2x)}{x^2}\right)' = \dfrac{2\cos(2x)x^2-\sin(2x)2x}{x^4}= \dfrac{2x\cos(2x)-2\sin(2x)}{x^3} $$

Answer 5

As an alternative by product rule $fg=f'g+fg'$ with

• $f(x)=\sin 2x \implies f'(x)=2\cos 2x$
• $g(x)=\frac{1}{x^2}\implies g'(x)=\frac{-2}{x^3}$

Answer 6

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.