How do you differentiate y= ln(3x^2-4) y=ln(3x24)?

2 Answers
Aug 1, 2017

y'-=dy/dx=(6x)/(3x^2-4)

Explanation:

For any y=ln[f(x)], y'-=dy/dx-=(f'(x))/(f(x))

y=ln(3x^2-4)

(3x^2-4)'=6xthereforey'=(6x)/(3x^2-4)

Aug 1, 2017

The derivative is =(6x)/(3x^2-4)

Explanation:

We need

(lnx)'=1/x

(x^n)'=nx^(n-1)

Here,

we have

y=ln(3x^2-4)

dy/dx=1/(3x^2-4)*(6x)=(6x)/(3x^2-4)