Question

How do you differentiate $y={{\sin }^{-1}}\left( 2x \right)$ ?

Answer 1

Hint: The inverse function of sine can be differentiated by the transposing the sine function to the left-hand side. Then we shall get a pure sine function in $y$ which would come out to be equal to $2x$. This can be differentiated with respect to $x$ variable. Since the derivative of sine function is the cosine function, therefore by using Pythagoras theorem, we will further find the value of $\cos y$.

Complete answer:
Any trigonometric function, $y={{\sin }^{-1}}\theta $ can be transformed as $\sin y=\theta $ by transposing the sine function to the left-hand side of the equation.
Thus, for the given equation, $y={{\sin }^{-1}}\left( 2x \right)$ the possible transformation is:
$\sin y=2x$
By Pythagorean understanding of trigonometry and its relation with right-angled triangles,
The value of sine function is equal to the perpendicular, $P$ of the right-angled triangle divided by its hypotenuse, $H$ which is further expressed as $\sin \theta =\dfrac{P}{H}$
Since, $\sin y=\dfrac{2x}{1}$, therefore, $P=2x$ and $H=1$
Similarly, the cosine function is given as the base, $B$ of the right-angled triangle divided by its hypotenuse,$H$ which is further expressed as $\cos \theta =\dfrac{B}{H}$
Using the Pythagoras theorem, ${{H}^{2}}={{P}^{2}}+{{B}^{2}}$,
We get ${{1}^{2}}={{\left( 2x \right)}^{2}}+{{B}^{2}}$
$\begin{align}
  & \Rightarrow 1=4{{x}^{2}}+{{B}^{2}} \\
 & \Rightarrow 1-4{{x}^{2}}={{B}^{2}} \\
\end{align}$
$\Rightarrow B=\sqrt{1-4{{x}^{2}}}$
Since,$\cos y=\dfrac{B}{H}$
Therefore, $\cos y=\dfrac{\sqrt{1-4{{x}^{2}}}}{1}=\sqrt{1-4{{x}^{2}}}$ …………………..equation (1)
Now, differentiating $\sin y=2x$ with respect to $x,$
Using chain rule of differentiation, $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$
{where,
$\dfrac{dy}{dx}=$ derivative of y with respect to x
$\dfrac{dy}{du}=$ derivative of y with respect to u
$\dfrac{du}{dx}=$ derivative of u with respect to x}
We get,
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \sin y \right)=\dfrac{d}{dx}\left( 2x \right) \\
 & \Rightarrow \cos y.\dfrac{dy}{dx}=2 \\
\end{align}$
$\therefore \dfrac{dy}{dx}=\dfrac{2}{\cos y}$
Substituting the value of $\cos y$ from equation (1), we get
$\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-4{{x}^{2}}}}$
Therefore, we get that the derivative of $y={{\sin }^{-1}}\left( 2x \right)$ is equal to $\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-4{{x}^{2}}}}$.

Note: The inverse of sine function is a common function used generally in daily mathematics. Therefore, it is more efficient to remember the derivative of the sine function. It is given as $\dfrac{d}{dx}\left[ {{\sin }^{-1}}\left( f\left( x \right) \right) \right]=\dfrac{1}{\sqrt{1-{{\left( f\left( x \right) \right)}^{2}}}}.{f}'\left( x \right)$ . It is derived by the chain rule of differentiation