Math whizzes~!! Is anyone able to solve this Q??

Question

Find y' by using logarithmic differentiation.

y=6(x^3+1)^2/(x^6)(e^-4x)

Answer 1

Assuming it was e^x - 4x

Rules of using ln:
(1) ln x*y = ln x + ln y
(2) ln x/y = ln x - ln y
(3) ln x^c = c * ln x

ln y = ln 6(x^3+1)^2 - ln (x^6)(e^x-4x) By (2)

ln y = ln 6 + ln (x^3+1)^2 - ln (x^6) - ln (e^x-4x) By (1) and (2)

ln y = ln 6 + 2 * ln (x^3+1) - ln (x^6) - ln (e^x-4x) By (3)

The derivative of the natural log is the derivative of the inside divided by the inside, i.e., the derivative of ln(g(x)) is g'(x)/g(x).

So taking the derivatives of both sides, you get

y ' / y = 0 + 2 (3x^2)/(x^3+1) - 6x^5/x^6 - (e^x-4)/(e^x-4x)

Multiply both sides by y to get

y ' = ( 2 (3x^2)/(x^3+1) - 6x^5/x^6 - (e^x-4)/(e^x-4x) ) * y

y ' = ( 2 (3x^2)/(x^3+1) - 6x^5/x^6 - (e^x-4)/(e^x-4x) ) * 6(x^3+1)^2/(x^6)(e^x-4x)

Answer 2

first rewrite as a product of two functions and we'll use the chain rule:

f(x) = g(x)h(x)
where g(x) = 6(x^3+1)^2
and h(x) = 1/[(x^6)*e^-4x]

f'(x) = g(x)h'(x) + g'(x)h(x)
= 6(x^3+1)^2h'(x) + 6*2*(x^3+1)*3*x^2h(x)

now, let h(x) = i(x)*j(x)
where i(x) = x^-6 and j(x) = e^4x
Use the same chain rule to differentiate h'(x) = i(x)j'(x) + i'(x)j(x)

Answer 3

1.39448 X 10^-15