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Question: What is the derivative of this function \(y={{\sin }^{-1}}{{x}^{2}}?\)...

What is the derivative of this function y=sin⁡−1x2?y={{\sin }^{-1}}{{x}^{2}}?

Explanation

Solution

We will use the formula given by ddxf(u)=dfdududx\dfrac{d}{dx}f\left( u \right)=\dfrac{df}{du}\dfrac{du}{dx} where ff is a function of uu and uu is a function of x.x. Also, we will use the trigonometric identity given by ddxsin⁡x=cos⁡x.\dfrac{d}{dx}\sin x=\cos x. Also, we will use the identity given by dxndx=nxn−1.\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}.

Complete step by step solution:
Let us consider the given function y=sin⁡−1x2.y={{\sin }^{-1}}{{x}^{2}}.
We are asked to find the derivative of the given inverse trigonometric function.
So, now, we can rearrange this function to a trigonometric function which is given by sin⁡y=x2.\sin y={{x}^{2}}.
We know that the derivative of the Sine function is the Cosine function and there is no change in the sign.
We can write this rule of differentiation for the concerned trigonometric function mathematically as ddxsin⁡x=cos⁡x.\dfrac{d}{dx}\sin x=\cos x.
Also, since yy is a function of x,x, we need to consider it separately.
We can write the formula as ddxf(u)=dfdududx\dfrac{d}{dx}f\left( u \right)=\dfrac{df}{du}\dfrac{du}{dx} where ff is a function of uu and uu is a function of x.x.
The same goes here. As we can see, here, u=yu=y and f(u)=sin⁡−1x2.f\left( u \right)={{\sin }^{-1}}{{x}^{2}}.
So, now let us use this rule in the rearranged form of the given equation.
We will get the derivative as ddxsin⁡y=dx2dx.\dfrac{d}{dx}\sin y=\dfrac{d{{x}^{2}}}{dx}.
Now, we will use the above identities on the left-hand side of the above equation and we will use the basic rule of differentiation dxndx=nxn−1\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}} on the right-hand side of the above equation.
We will get cos⁡ydydx=2x.\cos y\dfrac{dy}{dx}=2x.
Now, let us transpose the Cosine function to get dydx=2xcos⁡y.\dfrac{dy}{dx}=\dfrac{2x}{\cos y}.
We know that cos⁡x=1−sin⁡2x.\cos x=\sqrt{1-{{\sin }^{2}}x}.
So, when we use this, we will get dydx=2x1−sin⁡2y.\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{\sin }^{2}}y}}.
Now, we have sin⁡y=x2\sin y={{x}^{2}} and so, we will get dydx=2x1−(x2)2.\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{\left( {{x}^{2}} \right)}^{2}}}}.
And that is, dydx=2x1−x4.\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{x}^{4}}}}.
Hence the derivative is dydx=2x1−x4.\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{x}^{4}}}}.

Note: We know the Pythagorean identity sin⁡2x+cos⁡2x=1.{{\sin }^{2}}x+{{\cos }^{2}}x=1. So, now, if we transpose the Sine function, we will get cos⁡2x=1−sin⁡2x.{{\cos }^{2}}x=1-{{\sin }^{2}}x. And now from this, we will get cos⁡x=1−sin⁡2x.\cos x=\sqrt{1-{{\sin }^{2}}x}.