How do you find the derivative of y=ln(-(4x^4)/(x^3-3))^5?

1 Answer
Oct 22, 2017

y'=-(5(12-x^3))/(x(x^3-3))

Explanation:

The function in the general form:

y=ln(f(x))

then its derivative is

y'=1/f(x)*f'(x)

where f(x)=(color(red)g(x))^5

and its derivative is:

f'(x)=5(color(red)g(x))^4*g'(x)

where color(red)(g(x)=-(4x^4)/(x^3-3))

and its derivative is:

g'(x)=((-4*4x^3)(x^3-3)-(-4x^4)(3x^2))/(x^3-3)^2

=(-16x^6+48x^3+12x^6)/(x^3-3)^2

=(48x^3-4x^6)/(x^3-3)^2=(4x^3(12-x^3))/(x^3-3)^2

Finally, it is:

y'=1/(color(red)g(x))^5*5(color(red)g(x))^4*g'(x)

=1/(-(4x^4)/(x^3-3))^cancel5*5cancel((-(4x^4)/(x^3-3))^4)*(4x^3(12-x^3))/(x^3-3)^2

=-(5cancel((x^3-3)))/(cancel4x^cancel4)*(cancel(4x^3)(12-x^3))/(x^3-3)^cancel2

=-(5(12-x^3))/(x(x^3-3))