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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

  • 0
    You can also use that $\cos^2(x)=\frac 1 2 (1+\cos(2x))$ and differentiate that.2012-10-28

3 Answers 3

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

  • 0
    $\sin(2x)=2\sin(x)\cos(x)$2012-10-28
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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 $\cos(x)\cdot \sin(x) = \sin(s)$