Question

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

Answer 1

Hint: We will use the formula given by $\dfrac{d}{dx}f\left( u \right)=\dfrac{df}{du}\dfrac{du}{dx}$ where $f$ is a function of $u$ and $u$ is a function of $x.$ Also, we will use the trigonometric identity given by $\dfrac{d}{dx}\sin x=\cos x.$ Also, we will use the identity given by $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}.$

Complete step by step solution:
Let us consider the given function $y={{\sin }^{-1}}{{x}^{2}}.$
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 $\sin y={{x}^{2}}.$
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 $\dfrac{d}{dx}\sin x=\cos x.$
Also, since $y$ is a function of $x,$ we need to consider it separately.
We can write the formula as $\dfrac{d}{dx}f\left( u \right)=\dfrac{df}{du}\dfrac{du}{dx}$ where $f$ is a function of $u$ and $u$ is a function of $x.$
The same goes here. As we can see, here, $u=y$ and $f\left( u \right)={{\sin }^{-1}}{{x}^{2}}.$
So, now let us use this rule in the rearranged form of the given equation.
We will get the derivative as $\dfrac{d}{dx}\sin y=\dfrac{d{{x}^{2}}}{dx}.$
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 $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ on the right-hand side of the above equation.
We will get $\cos y\dfrac{dy}{dx}=2x.$
Now, let us transpose the Cosine function to get $\dfrac{dy}{dx}=\dfrac{2x}{\cos y}.$
We know that $\cos x=\sqrt{1-{{\sin }^{2}}x}.$
So, when we use this, we will get $\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{\sin }^{2}}y}}.$
Now, we have $\sin y={{x}^{2}}$ and so, we will get $\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{\left( {{x}^{2}} \right)}^{2}}}}.$
And that is, $\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{x}^{4}}}}.$
Hence the derivative is $\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{x}^{4}}}}.$

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