What is (\partial)/(\partial x)(6-(x^2+y^2)^{1/5}) ?

The answer to (\partial)/(\partial x)(6-(x^2+y^2)^{1/5}) is -(2x)/(5(x^2+y^2)^{4/5)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(6-(x^2+y^2)^{1/5})

\frac{\partial}{\partial x} (6 - (x^{2} + y^{2})^{\frac{1}{5}})

- \frac{2 x}{5 ( x ^{2} + y ^{2} ) ^{\frac{4}{5}}}

Solution steps

\frac{\partial}{\partial x} (6 - (x^{2} + y^{2})^{\frac{1}{5}})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (6) - \frac{\partial}{\partial x} ((x^{2} + y^{2})^{\frac{1}{5}})

\frac{\partial}{\partial x} (6) = 0

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{\frac{1}{5}}) = \frac{2 x}{5 ( x ^{2} + y ^{2} ) ^{\frac{4}{5}}}

= 0 - \frac{2 x}{5 ( x ^{2} + y ^{2} ) ^{\frac{4}{5}}}

Simplify

= - \frac{2 x}{5 ( x ^{2} + y ^{2} ) ^{\frac{4}{5}}}

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

y = x^{- 1}

y^{^{'}} = \frac{1}{x y}

f (x) = 8 x^{2} - 3, at x = 3

f (0) = \frac{cos ( x )}{x ^{2} - 1}

What is (\partial)/(\partial x)(6-(x^2+y^2)^{1/5}) ?
The answer to (\partial)/(\partial x)(6-(x^2+y^2)^{1/5}) is -(2x)/(5(x^2+y^2)^{4/5)}