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

  • $\begingroup$ I have typeset your question with LaTeX. Please double-check that I correctly transcribed it. $\endgroup$ – Neal Oct 28 '12 at 16:50
  • 4
    $\begingroup$ 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

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

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.