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

The derivative of 1-ln^2(x) is -(2ln(x))/x

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

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

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

- \frac{2 ln ( x )}{x}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (1) = 0

\frac{d}{d x} (ln^{2} (x)) = \frac{2 ln ( x )}{x}

= 0 - \frac{2 ln ( x )}{x}

Simplify

= - \frac{2 ln ( x )}{x}

\frac{\partial}{\partial x} (x^{2} - 7 x y + y)

\int csc^{2} (\frac{x}{5}) d x

\frac{\partial}{\partial x} (sin (x^{2} - 7 x y))

\int_{0}^{\infty} e^{- x} x d x

\frac{\partial}{\partial z} (sin (x))

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