Differentiate the function into the simplest form My question: $y=\sin^{2}(x)$
My attempt: Is $\sin^{2}x$ the same as $(\sin(x))^2$?
By rearranging the function I came up with the following.
$$ \begin{align}
u = \sin(x), \ & y=u^2 \\
\dfrac{du}{dx}= \cos(x), \ &\dfrac{dy}{dx}=2u \\
\end{align} $$
$$\dfrac{dy}{dx} = 2 \cos(x)$$
Is this correct? I am not sure if I'm going the correct way with the form. 
 A: Yes, $\sin^{2}x$ is the same as $(\sin(x))^2$.
Easiest way to calculate this derivative is by using the Chain rule. If
$$ f(x) = h(g(x))$$
then:
$$ f'(x) = h'(g(x))\cdot(g'(x)) $$
So in your case: $h(x) = x^2$ and $g(x)=\sin(x)$. That gives us:
$$\left( \sin^{2}x \right)' = 2\sin(x)\cdot (sin(x))' = 2\sin(x)\cos(x) $$
Which can be simplified by using following trigonometric property:
$$\sin(2x)=2\sin(x)\cos(x) $$
And in the end:
$$\left( \sin^{2}x \right)' = \sin(2x) $$
A: If we take $u = \sin(x),\; y=u^2$, then $$\dfrac{du}{dx} = \cos x\;\text{ and } \;\dfrac{dy}{du} = 2u$$
Note that you refer to $2u\,$ as $\;\dfrac{dy}{dx},\;$ which is incorrect, since we took the derivative of $y$ with respect to $u$, i.e. we computed $\dfrac{dy}{du}$.
Now what is true is that, by the chain rule, $$\dfrac{dy}{dx} = \dfrac{du}{dx}\cdot \dfrac{dy}{du}= \cos(x)\cdot 2u = \cos(x)\cdot 2\sin(x) = 2 \sin x \cos x$$
Note that we can simplify the result using the identity $$2\sin \alpha\cos \alpha = \sin(2\alpha)$$
A: It is not correct. Notice that $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx},$$ therefore $$\frac{dy}{dx} = (2u) \cdot (\cos (x)) = (2 \sin(x)) \cdot (\cos(x)) = 2 \sin (x) \cos (x) = \sin(2x).$$
