What is (\partial)/(\partial x)(8x(x^2+y^2)^{-1/2}) ?

The answer to (\partial)/(\partial x)(8x(x^2+y^2)^{-1/2}) is (8y^2)/((x^2+y^2)^{3/2)}

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(\partial)/(\partial x)(8x(x^2+y^2)^{-1/2})

\frac{\partial}{\partial x} (8 x (x^{2} + y^{2})^{- \frac{1}{2}})

\frac{8 y ^{2}}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}

Solution steps

\frac{\partial}{\partial x} (8 x (x^{2} + y^{2})^{- \frac{1}{2}})

Treat

y

as a constant

Take the constant out:

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

= 8 \frac{\partial}{\partial x} (x (x^{2} + y^{2})^{- \frac{1}{2}})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = (x^{2} + y^{2})^{- \frac{1}{2}}

= 8 (\frac{\partial}{\partial x} (x) (x^{2} + y^{2})^{- \frac{1}{2}} + \frac{\partial}{\partial x} ((x^{2} + y^{2})^{- \frac{1}{2}}) x)

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{- \frac{1}{2}}) = - \frac{x}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}

= 8 (1 \cdot (x^{2} + y^{2})^{- \frac{1}{2}} + (- \frac{x}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}) x)

Simplify

8 (1 \cdot (x^{2} + y^{2})^{- \frac{1}{2}} + (- \frac{x}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}) x) : \frac{8 y ^{2}}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}

= \frac{8 y ^{2}}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}}

What is (\partial)/(\partial x)(8x(x^2+y^2)^{-1/2}) ?
The answer to (\partial)/(\partial x)(8x(x^2+y^2)^{-1/2}) is (8y^2)/((x^2+y^2)^{3/2)}