What is the derivative of y = In (x^3 - 1)^4 times sqrt 3x - 1 / x^2 + 4

Question

I am getting confused about the finding of this derivative. Please show steps!

Answer 1

The Chain Rule says if y=f(g(x)) then y’=f’(g(x))g’(x). Let

f(((x³-1)⁴√(3x-1))/(x²+4))=ln(((x³-1)⁴√(3x-1))/(x²+4))

g(x)=((x³-1)⁴√(3x-1))/(x²+4)

Then, since (ln(x))’ = 1/x

f’(g(x)) = 1/((x³-1)⁴√(3x-1))/(x²+4)

Next, we must now determine g’(x)

Use the Quotient Rule which states that if g(x)=h(x)/k(x) then g’(x) = (k(x)h’(x) – h(x)k’(x))/(k(x))²

Here,

h(x)=(x³-1)⁴√(3x-1)

k(x)=(x²+4)

Using the Product Rule which states that if g(x)=h(x)k(x) then g’(x)=h’(x)k(x)+h(x)k’(x)

h’(x) = (4(x³-1)³*3x²)√(3x-1)+(x³-1)⁴*(1/2)/√(3x-1)*3

        = 12x²(x³-1)³√(3x-1)+(3/2)(x³-1)⁴/√(3x-1)

        = (12x²(x³-1)³+(3/2)(x³-1)⁴/(3x+1))/√(3x-1)

k’(x)=(x²+4)’=2x

From all this, we get

g’(x)=[√(3x-1)*(12x²(x³-1)³+(3/2)(x³-1)⁴/(3x+1))/√(3x-1)+√(3x-1)*2x]/(x²+4)²

So the final result is

(f(g(x)))’=[(1/((x³-1)⁴√(3x-1))/(x²+4))*[√(3x-1)*(12x²(x³-1)³+(3/2)(x³-1)⁴/(3x+1))/√(3x-1)+(x³-1)⁴√(3x-1)*2x]/(x²+4)²

              = [√(3x-1)*(12x²(x³-1)³+(3/2)(x³-1)⁴/(3x+1))/√(3x-1)+(x³-1)⁴√(3x-1)*2x]/(x²+4)²  / ((x³-1)⁴√(3x-1))

Answer 2

Since this function is a natural logarithmic function, you can use the Laws of Logarithm to rewrite the function.

So, it can first be written as y = ln(x³ – 1)⁴ + ln(3x – 1)^1/2 – ln(x² + 4).

This can be changed some more to y = 4ln(x³ – 1) + (1/2)ln(3x – 1) – ln(x² + 4).

From here, you can take the derivative of each term. You will need to use the derivative of y = ln x and the chain rule.

So you will get y' = 4·1/(x³ – 1)·(3x²) + (1/2)·1/(3x – 1)·(3) –1/(x² + 4)·(2x).

From here, you will simplify each term to finish finding the derivative.

y' = 12x²/(x³ – 1) +3/[2(3x – 1)] – 2x/(x² + 4)