How do you find the derivative of $y = ln (1 - (x^{2}))$ $y = ln(1-(x^2))$ ?

Question

11792 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-24T20:24:01

Recall that $\frac{d}{d x} ln (x) = \frac{1}{x}$ $d/dxln(x)=1/x$ .

So, by the chain rule, when we put a function inside the natural log function, we see that its derivative is

$d/dxln(f(x))=1/f(x)*f'(x)$

So here,

$dy/dx=d/dxln(1-x^2)=1/(1-x^2)[d/dx(1-x^2)]=(-2x)/(1-x^2)$

If you want, you can simplify the negative/minus signs:

$dy/dx=(2x)/(x^2-1)$

How do you find the derivative of y = ln(1-(x^2))y=ln(1−(x2))y = ln(1-(x^2))?