chiodaiokaC

Answered

2022-11-26

How do I differentiate ${\mathrm{sin}}^{2}x$?

I thought that because this is true:

${\mathrm{sin}}^{2}x=(\mathrm{sin}x{)}^{2},$

I could differentiate the expression like this:

$\frac{d}{dx}{\mathrm{sin}}^{2}x=2\mathrm{cos}x.$

But I am supposed to get

$\mathrm{sin}(2x)\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}2\mathrm{sin}x\mathrm{cos}x.$

Why am I wrong?

I thought that because this is true:

${\mathrm{sin}}^{2}x=(\mathrm{sin}x{)}^{2},$

I could differentiate the expression like this:

$\frac{d}{dx}{\mathrm{sin}}^{2}x=2\mathrm{cos}x.$

But I am supposed to get

$\mathrm{sin}(2x)\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}2\mathrm{sin}x\mathrm{cos}x.$

Why am I wrong?

Answer & Explanation

Alayna Phillips

Expert

2022-11-27Added 8 answers

You can only use the rule: $\frac{d}{dx}{x}^{n}=n{x}^{n-1}$ if x is just x and not a function. In the case for ${\mathrm{sin}}^{2}x$ then as you write it in the form $(\mathrm{sin}x{)}^{2}$ you can see that we can't use the power rule because it's not just an x term being raised to a constant power, it's a function being raised to a constant power. To differentiate ${\mathrm{sin}}^{2}x$ one must use the Chain Rule because $\mathrm{sin}x$ is a function of x within another function (the function that is squaring $\mathrm{sin}x$) ${x}^{2}$. The chain rule is: $\frac{d}{dx}f(g(x))={g}^{\prime}(x)\times ({f}^{\prime}(g(x))$ so you can apply that rule for this case with $f(x)={x}^{2}$ and $g(x)=\mathrm{sin}x$

ramirezherePYM

Expert

2022-11-28Added 1 answers

You are using the chain rule incorrectly:

$\frac{d}{dx}f(g(x))=\frac{df}{dx}(g(x))\frac{dg}{dx}(x)$

Now $f(x)={x}^{2}$ and $g(x)=\mathrm{sin}(x)$, therefore

$\frac{d}{dx}(\mathrm{sin}(x){)}^{2}=\stackrel{{f}^{\mathrm{\prime}}(g(x))}{\stackrel{\u23de}{2\mathrm{sin}(x)}}\cdot \stackrel{{g}^{\mathrm{\prime}}(x)}{\stackrel{\u23de}{\mathrm{cos}(x)}}$

And to explain your other point of confusion: $\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$ is a well-known trigonometric formula

$\frac{d}{dx}f(g(x))=\frac{df}{dx}(g(x))\frac{dg}{dx}(x)$

Now $f(x)={x}^{2}$ and $g(x)=\mathrm{sin}(x)$, therefore

$\frac{d}{dx}(\mathrm{sin}(x){)}^{2}=\stackrel{{f}^{\mathrm{\prime}}(g(x))}{\stackrel{\u23de}{2\mathrm{sin}(x)}}\cdot \stackrel{{g}^{\mathrm{\prime}}(x)}{\stackrel{\u23de}{\mathrm{cos}(x)}}$

And to explain your other point of confusion: $\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$ is a well-known trigonometric formula

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