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Popular >

derivative of y=ln(1-x^2)

Solution

derivative of

y = ln (1 - x^{2})

Solution

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

Solution steps

\frac{d}{d x} (ln (1 - x^{2}))

Apply the chain rule:

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

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

\frac{d}{d x} (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}}

Popular Examples

(\partial)/(\partial y)(2y^3-6y+x^2)

\frac{\partial}{\partial y} (2 y^{3} - 6 y + x^{2})

derivative of ln(5)derivative of

ln (5)

limit as x approaches+0+of 1/x-1/(x^2+x)

x \to + 0 + lim (\frac{1}{x} - \frac{1}{x ^{2} + x})

(\partial)/(\partial x)(e^{y/x}-y*e^{-x})

\frac{\partial}{\partial x} (e^{\frac{y}{x}} - y \cdot e^{- x})

limit as x approaches 0 of x^4sin(4/x)

x \to 0 lim (x^{4} sin (\frac{4}{x}))

Frequently Asked Questions (FAQ)

What is the derivative of y=ln(1-x^2) ?
The derivative of y=ln(1-x^2) is -(2x)/(1-x^2)
What is the first derivative of y=ln(1-x^2) ?
The first derivative of y=ln(1-x^2) is -(2x)/(1-x^2)