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How do you find the derivative of arcsinx+arccosx?

Answer
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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(x)=arcsinx+arccosx with respect to x.
We express the inverse function of sin and cos ratio in the form of arcsin(x)=sin1x and arccos(x)=cos1x.
Here we have a binary operation of addition for the main function is f(x)=arcsinx+arccosx and we convert the function into f(x)=sin1x+cos1x.
Here we take the functions where g(x)=sin1x and the other function is h(x)=cos1x.
We have f(x)=g(x)+h(x).
Differentiating f(x)=g(x)+h(x), we get
f(x)=ddx[f(x)]=ddx[g(x)+h(x)]=ddx[g(x)]+ddx[h(x)]=g(x)+h(x).
We know that differentiation of g(x)=sin1x is g(x)=11x2 and differentiation of h(x)=cos1x is h(x)=11x2.
We place the values of the differentiations and get
ddx[f(x)]=g(x)+h(x)=11x2+11x2
Simplifying we get 11x211x2=0.

Therefore, the differentiation of arcsinx+arccosx 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 sin1x+cos1x=π2 and then differentiate the function.
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