Using the differentiation rules and the derivative table, find the derivative of the function y=\arcsin(\sqrt{x})+\arctan(x^2)


y' = \frac{1}{2\sqrt{x(1 - x)}} + \frac{2x}{1 + x^4}

This is a sum of functions, so the derivative of y will be the sum of the derivatives of the two addends. To calculate them, we will use the derivation theorem for the composition of functions and the known derivatives of the inverse trigonometric functions.

  • Derivative of \arcsin(\sqrt{x})

    To derive \arcsin(\sqrt{x}) we note that it is a composition of functions with inner function g(x)=\sqrt{x} and outer function \arcsin(\cdot). Recall that the derivative of \arcsin(t) with respect to t is \frac{1}{\sqrt{1 - t^2}}, so from the Chain Rule we have

    \frac{d}{dx} \arcsin(\sqrt{x}) =\frac{1}{\sqrt{1 - (g(x))^2}} \cdot g'(x)=\frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx} (\sqrt{x})


    Since (\sqrt{x})^2 = x and the derivative of \sqrt{x} is \frac{1}{2\sqrt{x}} we find

    \frac{d}{dx} \arcsin(\sqrt{x}) = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x} \sqrt{1 - x}} = \frac{1}{2\sqrt{x(1 - x)}}


  • Derivative of \arctan(x^2)

    To derive \arctan(x^2), we use the Chain Rule again. In this case the inner function is g(x)=x^2, with derivative g'(x)=2x, and the outer function is \arctan(\cdot). Recalling that the derivative of \arctan(t) with respect to t is \frac{1}{1 + t^2}, from the Chain Rule we have

    \frac{d}{dx} \arctan(x^2) = \frac{1}{1 + (g(x))^2}\cdot g'(x)=\frac{1}{1 + (x^2)^2} \cdot 2x


    Since (x^2)^2 = x^4 we have

    \frac{d}{dx} \arctan(x^2) = \frac{1}{1 + x^4} \cdot 2x = \frac{2x}{1 + x^4}


  • Final derivative


    Let’s add the two derivatives obtained:


    • y' = \frac{d}{dx} \arcsin(\sqrt{x}) + \frac{d}{dx} \arctan(x^2)


    We get that the derivative of the function

    y = \arcsin(\sqrt{x}) + \arctan(x^2) è: y' = \frac{1}{2\sqrt{x(1 - x)}} + \frac{2x}{1 + x^4}