What is the derivative of sqrt(-ln(x^2+y)) ?

The derivative of sqrt(-ln(x^2+y)) is -(2x+\frac{dy)/(dx)}{2sqrt(-ln(x^2+y))(x^2+y)}

What is the first derivative of sqrt(-ln(x^2+y)) ?

The first derivative of sqrt(-ln(x^2+y)) is -(2x+\frac{dy)/(dx)}{2sqrt(-ln(x^2+y))(x^2+y)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 7 lim (x)

\frac{1}{s ^{2} + s + 1}

\int cos (x) (sec (x) tan (x)) d x

x \to - 2 lim (\frac{x + 2}{∣ x + 2 ∣})

x \to 0 lim (ln (x) sin (x))

What is the derivative of sqrt(-ln(x^2+y)) ?
The derivative of sqrt(-ln(x^2+y)) is -(2x+\frac{dy)/(dx)}{2sqrt(-ln(x^2+y))(x^2+y)}
What is the first derivative of sqrt(-ln(x^2+y)) ?
The first derivative of sqrt(-ln(x^2+y)) is -(2x+\frac{dy)/(dx)}{2sqrt(-ln(x^2+y))(x^2+y)}