Question

1. find the derivative of y with respect to the appropriate variable

(a) ( y = e^{-2x}ln(2x^{2}+2) ),
(b) ( y=(3x^{2}+x^{-2})cos3x ),
(c) ( y=sec x\tan x ).

Explanation:

Step1: Apply the product rule for (a)

The product rule is \((uv)^\prime = u^\prime v+uv^\prime\). Let \(u = e^{-2x}\), \(u^\prime=-2e^{-2x}\); \(v=\ln(2x^{2}+2)\), \(v^\prime=\frac{4x}{2x^{2}+2}=\frac{2x}{x^{2}+1}\). Then \(y^\prime=-2e^{-2x}\ln(2x^{2}+2)+e^{-2x}\frac{2x}{x^{2}+1}\)

Step2: Apply the product rule for (b)

Let \(u = 3x^{2}+x^{-2}\), \(u^\prime = 6x-2x^{-3}\); \(v=\cos3x\), \(v^\prime=-3\sin3x\). Then \(y^\prime=(6x - 2x^{-3})\cos3x-3(3x^{2}+x^{-2})\sin3x\)

Step3: Apply the product rule for (c)

Let \(u=\sec x\), \(u^\prime=\sec x\tan x\); \(v = \tan x\), \(v^\prime=\sec^{2}x\). Then \(y^\prime=\sec x\tan^{2}x+\sec^{3}x=\sec x(\tan^{2}x+\sec^{2}x)\)

Answer:

(a) \(y^\prime=-2e^{-2x}\ln(2x^{2}+2)+\frac{2xe^{-2x}}{x^{2}+1}\)
(b) \(y^\prime=(6x - 2x^{-3})\cos3x-3(3x^{2}+x^{-2})\sin3x\)
(c) \(y^\prime=\sec x(\tan^{2}x+\sec^{2}x)\)