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

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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.