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Popular Calculus >

d/(d{z)}({x}{y}e^{({x}^2+{z}^2-5)})

Solution

\frac{d}{d z} (x y e^{(x^{2} + z^{2} - 5)})

Solution

x y e^{x^{2} + z^{2} - 5} \cdot 2 z

Solution steps

\frac{d}{d z} (x y e^{(x^{2} + z^{2} - 5)})

Take the constant out:

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

= x y \frac{d}{d z} (e^{x^{2} + z^{2} - 5})

Apply the chain rule:

e^{x^{2} + z^{2} - 5} \frac{d}{d z} (x^{2} + z^{2} - 5)

= e^{x^{2} + z^{2} - 5} \frac{d}{d z} (x^{2} + z^{2} - 5)

\frac{d}{d z} (x^{2} + z^{2} - 5) = 2 z

= x y e^{x^{2} + z^{2} - 5} \cdot 2 z

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d er i v a t i v e 3 - x

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d er i v a t i v e y = x^{8 c o s (x)}

(\partial)/(\partial x)(5xye^{-x^2y})

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

limit as x approaches infinity of (0)^x

x \to \infty lim ((0)^{x})

derivative of (2/(x^3)^2)

\frac{d}{d x} ((\frac{2}{x ^{3}})^{2})

Frequently Asked Questions (FAQ)

What is the d/(d{z)}({x}{y}e^{({x}^2+{z}^2-5)}) ?
The d/(d{z)}({x}{y}e^{({x}^2+{z}^2-5)}) is {x}{y}e^{{x}^2+{z}^2-5}*2{z}
What is the first d/(d{z)}({x}{y}e^{({x}^2+{z}^2-5)}) ?
The first d/(d{z)}({x}{y}e^{({x}^2+{z}^2-5)}) is {x}{y}e^{{x}^2+{z}^2-5}*2{z}