What is (\partial)/(\partial x)(ln(9-x^2-y^2)) ?

The answer to (\partial)/(\partial x)(ln(9-x^2-y^2)) is -(2x)/(9-x^2-y^2)

Popular >

(\partial)/(\partial x)(ln(9-x^2-y^2))

\frac{\partial}{\partial x} (ln (9 - x^{2} - y^{2}))

- \frac{2 x}{9 - x ^{2} - y ^{2}}

Solution steps

\frac{\partial}{\partial x} (ln (9 - x^{2} - y^{2}))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{9 - x ^{2} - y ^{2}} \frac{\partial}{\partial x} (9 - x^{2} - y^{2})

= \frac{1}{9 - x ^{2} - y ^{2}} \frac{\partial}{\partial x} (9 - x^{2} - y^{2})

\frac{\partial}{\partial x} (9 - x^{2} - y^{2}) = - 2 x

= \frac{1}{9 - x ^{2} - y ^{2}} (- 2 x)

Simplify

\frac{1}{9 - x ^{2} - y ^{2}} (- 2 x) : - \frac{2 x}{9 - x ^{2} - y ^{2}}

= - \frac{2 x}{9 - x ^{2} - y ^{2}}

\frac{\partial}{\partial z} (z (y - yz)^{4})

\int_{0}^{5} e^{- x} d x

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 3

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

r^{2} - 16 x^{2}

What is (\partial)/(\partial x)(ln(9-x^2-y^2)) ?
The answer to (\partial)/(\partial x)(ln(9-x^2-y^2)) is -(2x)/(9-x^2-y^2)