Question

if ( f(x)=\tan^{-1}(sqrt{x}) ), find ( f^{prime}(x) ).

Explanation:

Step1: Apply the chain rule

The chain rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). Let \(u = \sqrt{x}=x^{\frac{1}{2}}\), so \(f(x)=\tan^{- 1}(u)\). The derivative of \(\tan^{-1}(u)\) with respect to \(u\) is \(\frac{1}{1 + u^{2}}\).

Step2: Differentiate \(u\) with respect to \(x\)

Differentiate \(u=x^{\frac{1}{2}}\) using the power rule \((x^n)^\prime=nx^{n - 1}\). So \(u^\prime=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\).

Step3: Substitute \(u\) and \(u^\prime\) back into the chain - rule formula

Substitute \(u = \sqrt{x}\) and \(u^\prime=\frac{1}{2\sqrt{x}}\) into \(f^\prime(x)=\frac{1}{1 + u^{2}}\cdot u^\prime\). We get \(f^\prime(x)=\frac{1}{1+(\sqrt{x})^{2}}\cdot\frac{1}{2\sqrt{x}}\).
Since \((\sqrt{x})^{2}=x\), then \(f^\prime(x)=\frac{1}{2\sqrt{x}(1 + x)}\).

Answer:

\(f^\prime(x)=\frac{1}{2\sqrt{x}(1 + x)}\)