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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

n = 1 \sum \infty \frac{n + 5}{5 ^{n}}

\int_{0}^{1} \frac{x ^{6}}{x ^{7} - 1} d x

\int \frac{3}{1 + x ^{2}} d x

y^{^{''}} + y = 4 x sin (x)

y^{^{''}} = - 4 \cdot y

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