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Question: What is the derivative of \(f\left( x \right)={{\cos }^{-1}}\left( x \right)\)?...

What is the derivative of f(x)=cos⁡−1(x)f\left( x \right)={{\cos }^{-1}}\left( x \right)?

Explanation

Solution

For solving this question, you should know about the differentiation of any function or differentiation of inverse trigonometric function because the value of the differentiation of trigonometric functions is also a trigonometric value but the differentiation of the inverse trigonometric functions are not trigonometric values. So, these are different from the differentiation of trigonometric functions in such a way.

Complete step by step answer:
In this question, we will obtain the derivative of cos⁡−1(x){{\cos }^{-1}}\left( x \right).
So, y=f(x)=cos⁡−1(x)→x=cos⁡yy=f\left( x \right)={{\cos }^{-1}}\left( x \right)\to x=\cos y from the definition of an inverse function.
Now differentiate both sides of x=cos⁡yx=\cos y, so we get,
ddxx=ddxcos⁡y\dfrac{d}{dx}x=\dfrac{d}{dx}\cos y
And by the properties of differentiation of inverse cosine function, we will get,
ddxx=−dydxsin⁡y ⇒1=−dydxsin⁡y \begin{aligned} & \dfrac{d}{dx}x=\dfrac{-dy}{dx}\sin y \\\ & \Rightarrow 1=\dfrac{-dy}{dx}\sin y \\\ \end{aligned}
So, we can write it as,
−1=dydxsin⁡y-1=\dfrac{dy}{dx}\sin y
Solving this for dydx\dfrac{dy}{dx}, we get,
dydx=−1sin⁡y………(1)\dfrac{dy}{dx}=\dfrac{-1}{\sin y}\ldots \ldots \ldots \left( 1 \right)
Since y=cos⁡−1(x)y={{\cos }^{-1}}\left( x \right), we can write here as, dydx=−1sin⁡(cos⁡−1x)\dfrac{dy}{dx}=-\dfrac{1}{\sin \left( {{\cos }^{-1}}x \right)} .
We know that sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1
So, in this replace the xx with cos⁡−1x{{\cos }^{-1}}x, so,
⇒(sin⁡2(cos⁡−1x))+(cos⁡2(cos⁡−1x))=1\Rightarrow \left( {{\sin }^{2}}\left( {{\cos }^{-1}}x \right) \right)+\left( {{\cos }^{2}}\left( {{\cos }^{-1}}x \right) \right)=1
We can further write it as,
cos⁡2(cos⁡−1x)=(cos⁡(cos⁡−1x))2=x2 sin⁡2(cos⁡−1x)+x2=1 ⇒sin⁡2(cos⁡−1x)=1−x2 \begin{aligned} & {{\cos }^{2}}\left( {{\cos }^{-1}}x \right)={{\left( \cos \left( {{\cos }^{-1}}x \right) \right)}^{2}}={{x}^{2}} \\\ & {{\sin }^{2}}\left( {{\cos }^{-1}}x \right)+{{x}^{2}}=1 \\\ & \Rightarrow {{\sin }^{2}}\left( {{\cos }^{-1}}x \right)=1-{{x}^{2}} \\\ \end{aligned}
Taking under root of both the sides, we get,
sin⁡(cos⁡−1x)=1−x2………(2)\sin \left( {{\cos }^{-1}}x \right)=\sqrt{1-{{x}^{2}}}\ldots \ldots \ldots \left( 2 \right)
Thus, we get,
dydx=−1sin⁡y\dfrac{dy}{dx}=-\dfrac{1}{\sin y}
By equation (1) and equation (2), we get,
dydx=−11−x2\dfrac{dy}{dx}=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}

So, the derivative of f(x)=cos⁡−1(x)f\left( x \right)={{\cos }^{-1}}\left( x \right) is −11−x2\dfrac{-1}{\sqrt{1-{{x}^{2}}}}.

Note: During solving these types of questions you must know about the formulas. All the basic formulas and other main formulas are used in this for solving this question. And you can also do it by making cos⁡(f(x))=x\cos \left( f\left( x \right) \right)=x and then taking the derivatives on both the sides and applying the same procedure after that.