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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~! ^_^

Ryan Evans
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  • You should say what $y$ represents (or if @A.Γ. 's edit is what you meant). – Arnaud Mortier Jun 21 '18 at 08:57
  • It could help to understand defining $q(y) = y^2$ and then using the chain rule on $\frac d{dx}q\bigl(\sin(x)\bigr)$. – Stefan Hante Jun 21 '18 at 08:58
  • sorry I don't understand either of these comments, unless they are addressing my original error in the title of writing $\frac{dy}{dx}$ instead of $\frac{d}{dx}$ – Ryan Evans Jun 21 '18 at 09:07

8 Answers8

6

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

Kenta S
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Chris2018
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We have that by the product rule, $$\begin{align}\frac{d}{dx}\sin^2(x)&=\frac{d}{dx}(\sin(x)\sin(x))\\&= (\frac{d}{dx}\sin(x))\sin(x)+\sin(x)(\frac{d}{dx}\sin(x))\\&= \cos(x)\sin(x)+\sin(x)\cos(x)\\&=2\sin(x)\cos(x).\end{align}$$

Robert Z
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Federico Fallucca
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    @RyanEvans It is correct. The only problem that it does not use the chain rule, so you cannot apply it to other problems in general. – A.Γ. Jun 21 '18 at 09:18
  • @A.Γ. ah.. good point.. – Ryan Evans Jun 21 '18 at 09:19
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    @RyanEvans If the middle step in Chris2006's answer is unclear then it is important to learn how the chain rule works and train it out. It is one of the major principles in derivation. – A.Γ. Jun 21 '18 at 09:21
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$$\dfrac{d(\sin^nx)}{dx}=\dfrac{d(\sin^nx)}{d(\sin x)}\cdot\dfrac{d(\sin x)}{dx}=?$$

See Chain rule

lab bhattacharjee
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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 $$ \color{blue}{2\sin x}\cdot\color{red}{\cos x}. $$

A.Γ.
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  • Apologies, I am unfamiliar with this ⟶ notation. – Ryan Evans Jun 21 '18 at 09:11
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    @RyanEvans It means "maps to", a convenient way to think of a mapping. $y=f(x)$ can be written as $x\to y$, that is, you take $x$ and construct $y$ of it. – A.Γ. Jun 21 '18 at 09:13
1

With bare hands:

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

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

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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

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

Rumplestillskin
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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

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

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

WuKong
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