Question

How do you find the derivative of $\dfrac{1}{2}\sin 2x$?

Answer 1

Hint: The given function $\dfrac{1}{2}\sin 2x$ is a composite function. We have to use the chain rule for the derivation. We derive the main function with respect to the secondary one. Then we take the derivation of the secondary function with respect to $x$. We take the multiplication of these two functions.

Complete step by step solution:
We differentiate the given function $f\left( x \right)=\dfrac{1}{2}\sin 2x$ with respect to $x$ using the chain rule.
Here we have a composite function where the main function is $g\left( x \right)=\sin x$ and the other function is $h\left( x \right)=2x$.
We have $goh\left( x \right)=g\left( 2x \right)=\sin 2x$. We take this as ours $f\left( x \right)=\dfrac{1}{2}\sin 2x$.
We need to find the value of $\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ \dfrac{1}{2}\sin 2x \right]$. We know $f\left( x \right)=\dfrac{1}{2}goh\left( x \right)$.
Differentiating $f\left( x \right)=\dfrac{1}{2}goh\left( x \right)$, we get
\[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ goh\left( x \right) \right]=\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}={{g}^{'}}\left[ h\left( x \right) \right]{{h}^{'}}\left( x \right)\].
The above-mentioned rule is the chain rule.
The chain rule allows us to differentiate with respect to the function $h\left( x \right)$ instead of $x$ and after that we need to take the differentiated form of $h\left( x \right)$ with respect to $x$.
For the function $f\left( x \right)=\dfrac{1}{2}\sin 2x$, we take differentiation of $f\left( x \right)=\dfrac{1}{2}\sin 2x$ with respect to the function $h\left( x \right)=2x$ instead of $x$ and after that we need to take the differentiated form of $h\left( x \right)=2x$ with respect to $x$.
The differentiation of $g\left( x \right)=\sin x$ is ${{g}^{'}}\left( x \right)=\cos x$ and differentiation of $h\left( x \right)=2x$ is \[{{h}^{'}}\left( x \right)=2\]. We apply the formula of \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\].
\[\Rightarrow \dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{d\left[ 2x \right]}\left[ \dfrac{1}{2}\sin 2x \right]\times \dfrac{d\left[ 2x \right]}{dx}\]
We place the values of the differentiations and get
\[\Rightarrow \dfrac{d}{dx}\left[ f\left( x \right) \right]=\left( \dfrac{1}{2}\cos 2x \right)\left[ 2 \right]=\cos 2x\]
Therefore, the differentiation of $\dfrac{1}{2}\sin 2x$ is \[\cos 2x\].

Note: We can also assume the secondary function as a new variable. For the given function $\dfrac{1}{2}\sin 2x$, we assume $z=2x$. Then we use the chain rule in the form of
\[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ g\left( z \right) \right]=\dfrac{d}{dz}\left[ g\left( z \right) \right]\times \dfrac{dz}{dx}={{g}^{'}}\left( z \right)\times {{z}^{'}}\].
We replace the value $h\left( x \right)=z=2x$.