What is (\partial)/(\partial x)((x^p+y^p)^{1/p}) ?

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

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(\partial)/(\partial x)((x^p+y^p)^{1/p})

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

x^{p - 1} (x^{p} + y^{p})^{\frac{1 - p}{p}}

Solution steps

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

Treat

p, y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

\frac{( x ^{p} + y ^{p} ) ^{\frac{1 - p}{p}}}{p} p x^{p - 1} : x^{p - 1} (x^{p} + y^{p})^{\frac{1 - p}{p}}

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

4 sin (2 x)

\frac{\partial}{\partial x} (3 x + 4 y + 4)

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\int \frac{11 x - 3}{( 2 x - 1 ) ( x - 3 )} d x

\frac{\partial}{\partial x} (\frac{5 x z}{x + y})

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