find the derivative of y with respect to x.\n\n( y=sec ^{-1}left(2 x^{2}+1\right) ), ( x>0 )\n\n( \frac{d…

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}})