What is (\partial)/(\partial z)(x^2+2y^2+3z^2) ?

The answer to (\partial)/(\partial z)(x^2+2y^2+3z^2) is 6z

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(\partial)/(\partial z)(x^2+2y^2+3z^2)

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

6 z

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial z} (2 y^{2}) = 0

\frac{\partial}{\partial z} (3 z^{2}) = 6 z

= 0 + 0 + 6 z

Simplify

= 6 z

\frac{\partial}{\partial x} (e^{- 5 x^{2} - y^{2}})

\int_{0}^{2} (e^{t} + e^{- t}) d t

\int_{e}^{e^{2}} ln (x) d x

n = 0 \sum \infty (- 1)^{2 n + 1}

\int \frac{1}{1 + 6 x} d x

What is (\partial)/(\partial z)(x^2+2y^2+3z^2) ?
The answer to (\partial)/(\partial z)(x^2+2y^2+3z^2) is 6z