Calculus Derivatives Help?

Question

I need to use quotient, chain and product rule with derivative of lnx to solve the following:

(√1+3x^2 )lnx^2

(3x^2+4) / (√1+x^2)

1/ (3x^2+2)^3

2lnx^3 / (√ 2-x^2)

(1+3x)^2 / (1+√x)^2

√ = Square Root of

/ = Divide by

Answer 1

Chain rule : (f o g)' = f' o g · g'

Product rule : (uv)' = u'v + v'u

Quotient rule : (f/g)' = (f'g - g'f)/(g^2)

1. d/dx [√(1+3x^2)ln(x^2)] =2 · d/dx [(1+3x^2)^(.5)ln(x)] because ln(x^2) = 2ln|x|. Using the chain and product rules:

1A : dy/dx = 2x^(-1)(1+3x^2)^(.5) - 6x(1+3x^2)^(-.5)

2. d/dx [3x^2 + 4]/[(1+x^2)^(.5)]. Using the quotient and product rules:

{6x√(1+x^2) + x[√(1+x^2)]^(-1)}/(1+x^2) then simplifying the numerator

2A: dy/dx = [6x√(1+x^2)]/(1+x^2) + x/[(1+x^2)√(1+x^2)]

3. d/dx [1/(3x^2 + 2)^3] = d/dx (3x^2 + 2)^(-3). Using the chain rule,

3A: dy/dx = -18x(3x^2 + 2)^(-4) = -18x/(3x^2 + 2)^4.

4. d/dx [2ln(x^3)/√(2 - x^2)] = 6 · d/dx [lnx/√(2 - x^2)]. Using the quotient and chain rules,

6 · [x^(-1)√(2-x^2) + xlnx√(2-x^2)^(-1)]/(2-x^2). Simplifying the numerator,

4A: dy/dx = [6√(2-x^2)]/([x(x - x^2)] + [6xlnx]/[(2-x^2)√(2-x^2)]

5. d/dx [(1+3x)^2/(1+√x)^2]. Using the chain and quotient rules,

[6x(1+3x)(1+√x)^2 - 2x^(-.5)(1+√x)]/(1+√x)^4. Grouping out the common (1+√x) and distributing you get

[(6x + 18x^2)(1+√x) - 2x^(-.5)]/(1+√x)^3. Simplifying the fraction,

5A: dy/dx = [(6x + 18x^2)/(1+√x)^2 - 2/[x(1+√x)^3].