What is (\partial)/(\partial x)(2-(x^2+y^2)^{1/7}) ?

The answer to (\partial)/(\partial x)(2-(x^2+y^2)^{1/7}) is -(2x)/(7(x^2+y^2)^{6/7)}

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(\partial)/(\partial x)(2-(x^2+y^2)^{1/7})

\frac{\partial}{\partial x} (2 - (x^{2} + y^{2})^{\frac{1}{7}})

- \frac{2 x}{7 ( x ^{2} + y ^{2} ) ^{\frac{6}{7}}}

Solution steps

\frac{\partial}{\partial x} (2 - (x^{2} + y^{2})^{\frac{1}{7}})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (2) - \frac{\partial}{\partial x} ((x^{2} + y^{2})^{\frac{1}{7}})

\frac{\partial}{\partial x} (2) = 0

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{\frac{1}{7}}) = \frac{2 x}{7 ( x ^{2} + y ^{2} ) ^{\frac{6}{7}}}

= 0 - \frac{2 x}{7 ( x ^{2} + y ^{2} ) ^{\frac{6}{7}}}

Simplify

= - \frac{2 x}{7 ( x ^{2} + y ^{2} ) ^{\frac{6}{7}}}

y = (x^{5} - 1)^{3}

x \to 0 lim (x sin (\frac{1}{∣ x ∣}))

- 2 e^{- x^{2}} x

\int \frac{1}{1 - x ^{3}} d x

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

What is (\partial)/(\partial x)(2-(x^2+y^2)^{1/7}) ?
The answer to (\partial)/(\partial x)(2-(x^2+y^2)^{1/7}) is -(2x)/(7(x^2+y^2)^{6/7)}