What is (\partial)/(\partial x)(x(x^2+y^2+z^2)^m) ?

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

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(\partial)/(\partial x)(x(x^2+y^2+z^2)^m)

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

(x^{2} + y^{2} + z^{2})^{m} + 2 x^{2} m (x^{2} + y^{2} + z^{2})^{m - 1}

Solution steps

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

Treat

y, z, m

as constants

Apply the Product Rule:

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

f = x, g = (x^{2} + y^{2} + z^{2})^{m}

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

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

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{m}) = m (x^{2} + y^{2} + z^{2})^{m - 1} \cdot 2 x

= 1 \cdot (x^{2} + y^{2} + z^{2})^{m} + m (x^{2} + y^{2} + z^{2})^{m - 1} \cdot 2 xx

Simplify

1 \cdot (x^{2} + y^{2} + z^{2})^{m} + m (x^{2} + y^{2} + z^{2})^{m - 1} \cdot 2 xx : (x^{2} + y^{2} + z^{2})^{m} + 2 x^{2} m (x^{2} + y^{2} + z^{2})^{m - 1}

= (x^{2} + y^{2} + z^{2})^{m} + 2 x^{2} m (x^{2} + y^{2} + z^{2})^{m - 1}

\frac{d}{d x} (\frac{1}{f ( x )})

\int_{1}^{- 1} (x - 1) (3 x + 2) d x

\frac{d}{d x} (\frac{2}{sin ( x ) + cos ( x )})

t \frac{d y}{d t} = t^{3} + 25 t^{3} y

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