Question

find the derivative of y with respect to x.

( y=sec ^{-1}left(2 x^{2}+1
ight) ), ( x>0 )

( \frac{d y}{d x}= )

Explanation:

Step1: Apply the chain rule

Let \(u = 2x^{2}+1\), then \(y=\sec^{-1}(u)\). The derivative of \(\sec^{-1}(u)\) with respect to \(u\) is \(\frac{1}{|u|\sqrt{u^{2}-1}}\) (for \(|u|> 1\)). Since \(x>0\), \(u = 2x^{2}+1>1\). So \(\frac{dy}{du}=\frac{1}{u\sqrt{u^{2}-1}}\) and \(\frac{du}{dx}=4x\).

Step2: Substitute \(u\) and simplify

By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Substitute \(u = 2x^{2}+1\) into \(\frac{dy}{du}\cdot\frac{du}{dx}\), we get \(\frac{4x}{(2x^{2}+1)\sqrt{(2x^{2}+1)^{2}-1}}\).
Simplify the denominator: \((2x^{2}+1)^{2}-1=(2x^{2}+1 - 1)(2x^{2}+1 + 1)=4x^{2}(x^{2}+1)\). So \(\sqrt{(2x^{2}+1)^{2}-1}=2x\sqrt{x^{2}+1}\).
Then \(\frac{dy}{dx}=\frac{4x}{(2x^{2}+1)\cdot2x\sqrt{x^{2}+1}}=\frac{2}{(2x^{2}+1)\sqrt{x^{2}+1}}\).

Answer:

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