Question

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

Answer 1

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

Explanation:

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)`