2. find the derivative of y with respect to the appropriate variable\n(a) ( y = e^{-2x}ln(2x^{2}+2) ),\n(b)…

2. find the derivative of y with respect to the appropriate variable\n(a) ( y = e^{-2x}ln(2x^{2}+2) ),\n(b) ( y=(3x^{2}+x^{-2})cos3x ),\n(c) ( y=sec x\tan x ).

2. find the derivative of y with respect to the appropriate variable\n(a) ( y = e^{-2x}ln(2x^{2}+2) ),\n(b) ( y=(3x^{2}+x^{-2})cos3x ),\n(c) ( y=sec x\tan x ).

Answer

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))