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
$\endgroup$ 33 Answers
$\begingroup$$2\cos (x)(-\sin(x))=-2\cos(x)\sin(x)=-\sin(2x)$
$\endgroup$ 2 $\begingroup$Chain Rule
To differentiate $y = \cos^2x$ with respect to $x$, one must apply the chain rule as shown:
$$ \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)$$
$\endgroup$ $\begingroup$You did a mistake there - you misuse the trigonometric identity. Look at
It's $$2 \cdot \cos(x)\cdot \sin(x) = \sin(2x)$$ Not $$\cos(x)\cdot \sin(x) = \sin(s)$$
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