Question

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

Answer 1

Hint: 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 $x$ 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 $f\left( x \right)=\arcsin x+\arccos x$ with respect to $x$.
We express the inverse function of sin and cos ratio in the form of $\arcsin \left( x \right)={{\sin }^{-1}}x$ and $\arccos \left( x \right)={{\cos }^{-1}}x$.
Here we have a binary operation of addition for the main function is $f\left( x \right)=\arcsin x+\arccos x$ and we convert the function into $f\left( x \right)={{\sin }^{-1}}x+{{\cos }^{-1}}x$.
Here we take the functions where $g\left( x \right)={{\sin }^{-1}}x$ and the other function is $h\left( x \right)={{\cos }^{-1}}x$.
We have $f\left( x \right)=g\left( x \right)+h\left( x \right)$.
Differentiating $f\left( x \right)=g\left( x \right)+h\left( x \right)$, we get
\[{{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ g\left( x \right)+h\left( x \right) \right]=\dfrac{d}{dx}\left[ g\left( x \right) \right]+\dfrac{d}{dx}\left[ h\left( x \right) \right]={{g}^{'}}\left( x \right)+{{h}^{'}}\left( x \right)\].
We know that differentiation of $g\left( x \right)={{\sin }^{-1}}x$ is ${{g}^{'}}\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ and differentiation of $h\left( x \right)={{\cos }^{-1}}x$ is \[{{h}^{'}}\left( x \right)=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}\].
We place the values of the differentiations and get
\[\Rightarrow \dfrac{d}{dx}\left[ f\left( x \right) \right]={{g}^{'}}\left( x \right)+{{h}^{'}}\left( x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}+\dfrac{-1}{\sqrt{1-{{x}^{2}}}}\]
Simplifying we get \[\dfrac{1}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}}=0\].

Therefore, the differentiation of $\arcsin x+\arccos x$ 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 ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ and then differentiate the function.