What is the (ln(1-x^2))^' ?

The (ln(1-x^2))^' is -(2x)/(1-x^2)

What is the first (ln(1-x^2))^' ?

The first (ln(1-x^2))^' is -(2x)/(1-x^2)

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(ln (1 - x^{2}))^{^{'}}

- \frac{2 x}{1 - x ^{2}}

Solution steps

(ln (1 - x^{2}))^{'}

Apply the chain rule:

\frac{1}{1 - x ^{2}} (1 - x^{2})^{'}

= \frac{1}{1 - x ^{2}} (1 - x^{2})^{'}

(1 - x^{2})^{'} = - 2 x

= \frac{1}{1 - x ^{2}} (- 2 x)

Simplify

\frac{1}{1 - x ^{2}} (- 2 x) : - \frac{2 x}{1 - x ^{2}}

= - \frac{2 x}{1 - x ^{2}}

x y^{2} y^{^{'}} = y^{3} - 2 x^{3}

\frac{d}{d x} (\frac{1}{4 + 3 x})

\int_{1}^{2} x^{5} + \frac{1}{4 x ^{5}} d x

\int \frac{1}{( x ^{2} + 1 ) ^{3}} d x

\frac{\partial}{\partial x} (ln (x^{2} + x y))