How to evaluate $\frac{d}{dx}\sin^2x$

Question

I'm having trouble understanding how $\frac{d}{dx}\sin^2x =2\sin(x)\cos(x)$.

Please show as many steps of the proof as necessary so that I can apply this to other problems.

Thank you for your time~! ^_^

Answer 1

$\frac{d}{dx}\sin^2x= 2\sin(x)(\sin(x))'= 2\sin(x)\cos(x)$

Answer 2

$$\dfrac{d(\sin^nx)}{dx}=\dfrac{d(\sin^nx)}{d(\sin x)}\cdot\dfrac{d(\sin x)}{dx}=?$$

See Chain rule

Answer 3

We have that by the product rule, $$\begin{align}\frac{d}{dx}\sin^2(x)&=\frac{d}{dx}(\sin(x)\sin(x))\\&= (\frac{d}{dx}\sin(x))\sin(x)+\sin(x)(\frac{d}{dx}\sin(x))\\&= \cos(x)\sin(x)+\sin(x)\cos(x)\\&=2\sin(x)\cos(x).\end{align}$$

Answer 4

You have a composite function $$ x\quad\color{red}{\longrightarrow}\quad u=\sin x\quad\color{blue}{\longrightarrow}\quad y=u^2. $$

1. The derivative of $u\color{blue}\to u^2$ is $2u$, but $u=\sin x$, then it is $\color{blue}{2\sin x}$.
2. The derivative of $x\color{red}\to\sin x$ is $\color{red}{\cos x}$.

The chain rule says: take their product $$ \color{blue}{2\sin x}\cdot\color{red}{\cos x}. $$

Answer 5

With bare hands:

$$\sin^2x=\frac{1-\cos2x}2$$ so that

$$(\sin^2x)'=-\frac12(-2\sin2x)=\sin2x=2\sin x\cos x.$$

Answer 6

The derivative of a function $(f(x))^n$ is given by

$$ ((f(x))^n)' = \frac{d (f(x))^n }{ d f(x)} \frac{d f(x)}{dx} = n f(x)^{n-1} f'(x), $$

by the chain rule.

Hence for your example $(f(x))^2 = (\sin(x))^2$ and the derivative is

$$\left((\sin(x))^2 \right)' = 2 \sin(x) \cdot (\sin(x))' = 2 \sin(x) \cos(x). $$

Answer 7

Using the definition of the derivative and the general theorem $\lim(fg)=(\lim f)(\lim g)$ provided $\lim f$ and $\lim g$ both exist, we have

$$\begin{align} {d\over dx}\sin^2x &=\lim_{y\to x}{\sin^2x-\sin^2y\over x-y}\\ &=\left(\lim_{y\to x}(\sin x+\sin y)\right)\left(\lim_{y\to x}{\sin x-\sin y\over x-y} \right)\\ &=(\sin x+\sin x)\left({d\over dx}\sin x\right)\\ &=2\sin x\cos x \end{align}$$

Answer 8

$$\frac{{\rm d}(\sin^2 x)}{{\rm d}x}=\frac{{\rm d}(\sin^2 x)}{{\rm d}(\sin x)}\cdot \frac{{\rm d}(\sin x)}{{\rm d}x}=2\sin x \cos x.$$