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Question: How do you find the derivative of \(\arcsin x+\arccos x\)?...

How do you find the derivative of arcsin⁡x+arccos⁡x\arcsin x+\arccos x?

Explanation

Solution

We first define how differentiation works for composite function in a binary operation of addition. We take differentiation of the functions separately with respect to xx and apply the same operation on them. We take the operation’s answer as the final solution.

Complete step by step solution:
We differentiate the given function f(x)=arcsin⁡x+arccos⁡xf\left( x \right)=\arcsin x+\arccos x with respect to xx.
We express the inverse function of sin and cos ratio in the form of arcsin⁡(x)=sin⁡−1x\arcsin \left( x \right)={{\sin }^{-1}}x and arccos⁡(x)=cos⁡−1x\arccos \left( x \right)={{\cos }^{-1}}x.
Here we have a binary operation of addition for the main function is f(x)=arcsin⁡x+arccos⁡xf\left( x \right)=\arcsin x+\arccos x and we convert the function into f(x)=sin⁡−1x+cos⁡−1xf\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x.
Here we take the functions where g(x)=sin⁡−1xg\left( x \right)={{\sin }^{-1}}x and the other function is h(x)=cos⁡−1xh\left( x \right)={{\cos }^{-1}}x.
We have f(x)=g(x)+h(x)f\left( x \right)=g\left( x \right)+h\left( x \right).
Differentiating f(x)=g(x)+h(x)f\left( x \right)=g\left( x \right)+h\left( x \right), we get
f′(x)=ddx[f(x)]=ddx[g(x)+h(x)]=ddx[g(x)]+ddx[h(x)]=g′(x)+h′(x){{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ g\left( x \right)+h\left( x \right) \right]=\dfrac{d}{dx}\left[ g\left( x \right) \right]+\dfrac{d}{dx}\left[ h\left( x \right) \right]={{g}^{'}}\left( x \right)+{{h}^{'}}\left( x \right).
We know that differentiation of g(x)=sin⁡−1xg\left( x \right)={{\sin }^{-1}}x is g′(x)=11−x2{{g}^{'}}\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}} and differentiation of h(x)=cos⁡−1xh\left( x \right)={{\cos }^{-1}}x is h′(x)=−11−x2{{h}^{'}}\left( x \right)=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}.
We place the values of the differentiations and get
⇒ddx[f(x)]=g′(x)+h′(x)=11−x2+−11−x2\Rightarrow \dfrac{d}{dx}\left[ f\left( x \right) \right]={{g}^{'}}\left( x \right)+{{h}^{'}}\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}+\dfrac{-1}{\sqrt{1-{{x}^{2}}}}
Simplifying we get 11−x2−11−x2=0\dfrac{1}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}}=0.

Therefore, the differentiation of arcsin⁡x+arccos⁡x\arcsin x+\arccos x is 0.

Note: Each of the six basic trigonometric functions have corresponding inverse functions when appropriate restrictions are placed on the domain of the original functions. All the inverse trigonometric functions have derivatives when appropriate restrictions are placed on the domain of the original functions.
For simplification we can also use the identity formula of sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} and then differentiate the function.