Question

How do you find the derivative of $y=\dfrac{1}{2\sin \left( 2x \right)}$?

Answer 1

As we all know if x and y are real numbers, and if the graph of f is plotted against x, then the derivative of the function is the slope of this graph at each point. There are many rules to find the derivative of functions. To find the derivative of above equation$y=\dfrac {1} {2\sin \left (2x \right)}$ with respect to $x$where$x$is independent variable and$y$ is dependent variable if we do any change in $x$ then $y$ also changes with the same rate. we will use quotient rule of differentiation which states that If two differentiable function $f\left( x \right)$ and $g\left( x \right)$are in a ration form like $\dfrac{f\left( x \right)}{g\left( x \right)}$then the derivative of the equation is$\dfrac{d\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)}{dy}=\dfrac{f'{\ }\left( x \right)g\left( x \right)-g'{\ }\left( x \right)f\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$ let's take $f\left( x \right)=1$ and $g\left( x \right)=\sin \left( 2x \right)$

Complete step by step answer:
The given equation is
$y=\dfrac{1}{2\sin \left( 2x \right)}$
Now by using quotient rule $\dfrac{d\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)}{dy}=\dfrac{f'{\ }\left( x \right)g\left( x \right)-g'{\ }\left( x \right)f\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$ on above given equation, where $f'{\ }\left( x \right)$is the differentiation of $f\left( x \right)$ and $f\left( x \right)$ is equal to 1
And $g'{\ }\left( x \right)$is the differentiation of $g\left( x \right)$ and $g\left( x \right)$ is equal to $\sin 2x$
Now,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\left( \dfrac{\left( 1 \right)'{\ }\sin 2x-\left( \sin 2x \right)'{\ }}{{{\sin }^{2}}2x} \right)..........\left( 1 \right)$
Now differentiation of $\dfrac{d\left( 1 \right)}{dx}=0$ and differentiation of $\dfrac{d\left( \sin 2x \right)}{dx}=2\cos 2x$.
Now put these above values in equation (1), we get
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\left( \dfrac{0\cdot \sin 2x-2\cos 2x}{{{\sin }^{2}}2x} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\cdot \left( \dfrac{-2\cos 2x}{{{\sin }^{2}}2x} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-\cos 2x}{{{\sin }^{2}}2x} \\\ \end{aligned}$
Hence by applying simply the quotient rule $\dfrac{d\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)}{dy}=\dfrac{f'{\ }\left( x \right)g\left( x \right)-g'{\ }\left( x \right)f\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$the derivative of $y=\dfrac{1}{2\sin \left( 2x \right)}$ is$\dfrac{-\cos 2x}{{{\sin }^{2}}2x}$.
Note: Here we simply applied the differentiation just by using quotient rule $\dfrac{d\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)}{dy}=\dfrac{f'{\ }\left( x \right)g\left( x \right)-g'{\ }\left( x \right)f\left( x \right)}{{{\left( g\left( x \right) \right)}^{2}}}$ .we can go wrong by converting $\sin 2x=2\sin x\cos x$ because its became lengthy and need more attention that's why we don’t use it. So carefully solve the question. To find the derivative of any equation we should know the differentiation of functions. Then it became too easy to apply differentiation rules. Here we have a function sin2x whose differentiation is 2cos2x.