# Differentiating cosine squared of x

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

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

## 3 Answers

$2\cos (x)(-\sin(x))=-2\cos(x)\sin(x)=-\sin(2x)$

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

## 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:

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

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 $$\cos(x)\cdot \sin(x) = \sin(s)$$