Question

1. find the derivative of y.

$y=sec^{-1}(2x + 1)$

Explanation:

Step1: Apply the chain rule

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

Step2: Find the derivative of \(u\)

Since \(u = 2x+1\), then \(u^\prime=\frac{d}{dx}(2x + 1)=2\).

Step3: Combine using the chain rule

Substitute \(u = 2x + 1\) and \(u^\prime=2\) into the chain - rule formula. \(y^\prime=\frac{1}{|2x + 1|\sqrt{(2x + 1)^{2}-1}}\cdot2=\frac{2}{|2x + 1|\sqrt{4x^{2}+4x}}\).

Answer:

\(y^\prime=\frac{2}{|2x + 1|\sqrt{4x^{2}+4x}}\)