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

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Answer 1 · 2017-08-01T17:54:51

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

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)#

Answer 2 · 2017-08-01T17:56:13

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

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)#