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

$y=ln(3x^2-4)$

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

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