What is the derivative of # ln[-30(x^3-2x+e^x)^5]#?

Question

Christopher Allers · Accepted Answer

# d/dxln{-30(x^3-3x+e^x)^5}=(5(3x^2-3+e^x))/(x^3-3x+e^x) #

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If # y=f(x) # then # f'(x)=dy/dx=dy/(du)(du)/dx #

I was taught to remember that the differential can be treated like a fraction and that the "#dx#'s" of a common variable will "cancel" (It is important to realise that #dy/dx# isn't a fraction but an operator that acts on a function, there is no such thing as "#dx#" or "#dy#" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

# dy/dx = dy/(dv)(dv)/(du)(du)/dx # etc, or # (dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx) #

So If # y=ln{-30(x^3-3x+e^x)^5} #, Then we can rewrite using the laws of logarithms as follows;: # y=ln(-30) + ln(x^3-3x+e^x)^5 # # :. y=ln(-30) + 5ln(x^3-3x+e^x) #

# { ("Let "u=x^3-3x+e^x, => , (du)/dx=3x^2-3+e^x), ("Then "y=ln(-30) + 5lnu, =>, dy/(du)=5/u ) :}#

Using # dy/dx=(dy/(du))((du)/dx) # we get:

# dy/dx=(5/u)(3x^2-3+e^x) # # :. dy/dx=(5(3x^2-3+e^x))/(x^3-3x+e^x) #

Emily Ammerman · Answer

undefined To find the derivative of ln[-30(x^3-2x+e^x)^5], apply the chain rule. The derivative is:

d/dx [ln(u)] = (1/u) * du/dx

Where u = -30(x^3 - 2x + e^x)^5

So, the derivative is:

d/dx [ln(-30(x^3-2x+e^x)^5)] = (1/[-30(x^3-2x+e^x)^5]) * d/dx [-30(x^3-2x+e^x)^5]

Then apply the chain rule to differentiate the expression inside the logarithm. The derivative of (-30(x^3 - 2x + e^x)^5) is:

d/dx [-30(x^3-2x+e^x)^5] = -150(x^3-2x+e^x)^4 * (3x^2 - 2 + e^x)

So, putting it all together:

d/dx [ln[-30(x^3-2x+e^x)^5]] = (1/[-30(x^3-2x+e^x)^5]) * [-150(x^3-2x+e^x)^4 * (3x^2 - 2 + e^x)]