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

Question

William Abdalla · Accepted Answer

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

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))#

Samantha Adkison · Answer

undefined To find the derivative of $ y = \ln\left(-\frac{4x^4}{x^3 - 3}\right)^5 $, you can use the chain rule and the derivative of the natural logarithm function.

First, differentiate the outer function $ u^n $ with respect to $ u $, where $ u = \ln\left(-\frac{4x^4}{x^3 - 3}\right) $, and then multiply by the derivative of the inner function.

The derivative of $ u^n $ with respect to $ u $ is $ n \cdot u^{n-1} $.

The derivative of $ \ln(u) $ with respect to $ u $ is $ \frac{1}{u} $.

The derivative of $ -\frac{4x^4}{x^3 - 3} $ with respect to $ x $ is found using the quotient rule.

Let's denote $ f(x) = -4x^4 $ and $ g(x) = x^3 - 3 $.

Then, $ \frac{d}{dx} \left(-\frac{4x^4}{x^3 - 3}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} $.

Now, you can apply these steps to find the derivative of $ y $.