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

Note : In the above problem we have solved the derivative of inverse trigonometric function. In (3) the formation of $\dfrac{1}{2\sqrt{x}}$ is due to function in a function. In this case we have to find an internal derivative. Further solving for $\dfrac{dy}{dx}$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.