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I am trying to differentiate with respect to $x$, $y = \cos^2{x}$

Using the chain rule and my working out is this:

\begin{align*}\frac{dy}{dx} &= 2 \cos(x)(-\sin(x)) \\ &= -2 \sin x \end{align*} I am not sure how to get to the correct answer of $-\sin{(2x)}$.

Should I be using the chain rule or maybe the product rule?

Please help, Thanks in advance

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  • $\begingroup$ I have typeset your question with LaTeX. Please double-check that I correctly transcribed it. $\endgroup$ – Neal Oct 28 '12 at 16:50
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    $\begingroup$ Your differentiation is correct. $2\cos x(-\sin x) \ne -2 \sin x$ but $2\cos x(-\sin x)=-2\sin{x}\cos{x}=-\sin{2x}$ $\endgroup$ – M. Strochyk Oct 28 '12 at 16:51
  • $\begingroup$ You can also use that $\cos^2(x)=\frac 1 2 (1+\cos(2x))$ and differentiate that. $\endgroup$ – Mark Bennet Oct 28 '12 at 16:54
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$2\cos (x)(-\sin(x))=-2\cos(x)\sin(x)=-\sin(2x)$

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  • $\begingroup$ Thanks for clearing that up, what trig identity have you used.. thanks $\endgroup$ – user866190 Oct 28 '12 at 18:39
  • $\begingroup$ $\sin(2x)=2\sin(x)\cos(x)$ $\endgroup$ – Mark Bennet Oct 28 '12 at 19:50
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Chain Rule

To differentiate $y = \cos^2x$ with respect to $x$, one must apply the chain rule as shown:

$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ Firstly, $ \,\,let \,\,\, u = \cos x \,\,$
One can then differentiate this with respect to $x$ such that $$\frac{du}{dx} = -sinx$$ Then, $ \,\,let \,\,\, y = u^2$
Differentiate $y$ with respect to $u$ such that $\frac{dy}{du} = 2u$

Next, one can substitute $u$ back in to make $$\frac{dy}{du} = 2\cos x$$

Thus, $$\frac{dy}{dx} = -2\sin x \cdot \cos x$$

Double Angle Formula Simplification

Using the formula: $$\sin(2u) = 2\sin u\cos u$$
We can simplify to:

$$\frac{dy}{dx} = -\sin(2x)$$

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You did a mistake there - you misuse the trigonometric identity. Look at http://www.sosmath.com/trig/Trig5/trig5/trig5.html

It's $$2 \cdot \cos(x)\cdot \sin(x) = \sin(2x)$$ Not $$\cos(x)\cdot \sin(x) = \sin(s)$$

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