What is (\partial)/(\partial y)(9sqrt(x^2+y^2)) ?

The answer to (\partial)/(\partial y)(9sqrt(x^2+y^2)) is (9y)/(sqrt(x^2+y^2))

Popular >

(\partial)/(\partial y)(9sqrt(x^2+y^2))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

= 9 \cdot \frac{1}{2 x ^{2} + y ^{2}} \cdot 2 y

Simplify

9 \cdot \frac{1}{2 x ^{2} + y ^{2}} \cdot 2 y : \frac{9 y}{x ^{2} + y ^{2}}

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

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

\frac{d y}{d t} = 2 y - t^{2}

\int_{0}^{2} x 6 - x^{2} d x

\int \frac{x ^{2}}{4 - x ^{3}} d x

x = 1 \sum \infty \frac{x}{x ^{2} + 1}

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