Question

Find the derivative of the following
$$y={\tan }^{-1}\sqrt{x}$$

Answer 1

Hint: To solve the above problem we have to know the basic derivatives of $${\tan }^{-1}x$$and $$\sqrt{x}$$. After writing the derivatives rewrite the equation with the derivatives of the function.
$${\displaystyle \frac{d}{dx}}\left({\tan }^{-1}x\right)={\displaystyle \frac{1}{1+x^2}}$$, $${\displaystyle \frac{d}{dx}}\sqrt{x}={\displaystyle \frac{1}{2\sqrt{x}}}$$. We can see one function is inside another we have to find internal derivatives.

Complete step-by-step answer:
The composite function rule shows us a quicker way. If f(x) = h(g(x)) then f (x) = h (g(x)) × g (x). In words: differentiate the 'outside' function, and then multiply by the derivative of the 'inside' function. ... The composite function rule tells us that f (x) = 17(x2 + 1)16 × 2x.

$$y={\tan }^{-1}\sqrt{x}$$. . . . . . . . . . . . . . . . . . . . . (a)
$${\displaystyle \frac{d}{dx}}\left({\tan }^{-1}x\right)={\displaystyle \frac{1}{1+x^2}}$$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
$${\displaystyle \frac{d}{dx}}\sqrt{x}={\displaystyle \frac{1}{2\sqrt{x}}}$$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substituting (1) and (2) as derivatives we get,
Therefore derivative of the given function is,
$$y^1={\displaystyle \frac{d}{dx}}\left({\tan }^{-1}\sqrt{x}\right)$$
We know the derivative of $${\tan }^{-1}x$$and $$\sqrt{x}$$. By writing the derivatives we get,
Further solving we get the derivative of the function as
$$y^1={\displaystyle \frac{1}{1+{\left(\sqrt{x}\right)}^2}}{\displaystyle \frac{d}{dx}}\left(\sqrt{x}\right)$$. . . . . . . . . . . . . . . . . . . (3)
By solving we get,
$$y^1={\displaystyle \frac{1}{1+x}}\times {\displaystyle \frac{1}{2\sqrt{x}}}$$
Multiplying $$2\sqrt{x}$$ with (1+x) and expanding we get,
$$y^1={\displaystyle \frac{1}{2\sqrt{x}+2x\sqrt{x}}}$$
We know that $$\sqrt{x}$$ can be written as $$x^{{\displaystyle \frac{1}{2}}}$$.
By expressing $$\sqrt{x}$$ as $$x^{{\displaystyle \frac{1}{2}}}$$ we get,
$$y^1={\displaystyle \frac{1}{2{\left(x\right)}^{{\displaystyle \frac{1}{2}}}+2x\cdot {\left(x\right)}^{{\displaystyle \frac{1}{2}}}}}$$
Applying the rule $$x\cdot x^{{\displaystyle \frac{1}{2}}}=x^{{\displaystyle \frac{3}{2}}}$$ we get,
$$y^1={\displaystyle \frac{1}{2{\left(x\right)}^{{\displaystyle \frac{1}{2}}}+2{\left(x\right)}^{{\displaystyle \frac{3}{2}}}}}$$

Note: In the above problem we have solved the derivative of inverse trigonometric function. In (3) the formation of $${\displaystyle \frac{1}{2\sqrt{x}}}$$ is due to function in a function. In this case we have to find an internal derivative. Further solving for $${\displaystyle \frac{dy}{dx}}$$made us towards a solution. If we are doing derivative means we are finding the slope of a function. Care should be taken while doing calculations.