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Popular >

d/(d{y)}(({z}(y,x))(e^{{x}{z}(y,x)}+{x}{y}))

Solution

\frac{d}{d y} ((z (y, x)) (e^{x z (y, x)} + x y))

Solution

z (y, x) (0 + x)

Solution steps

\frac{d}{d y} ((z (y, x)) (e^{x z (y, x)} + x y))

Take the constant out:

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

= z (y, x) \frac{d}{d y} (e^{x z (y, x)} + x y)

Apply the Sum/Difference Rule:

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

= z (y, x) (\frac{d}{d y} (e^{x z (y, x)}) + \frac{d}{d y} (x y))

\frac{d}{d y} (e^{x z (y, x)}) = 0

\frac{d}{d y} (x y) = x

= z (y, x) (0 + x)

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Frequently Asked Questions (FAQ)

What is the d/(d{y)}(({z}(y,x))(e^{{x}{z}(y,x)}+{x}{y})) ?
The d/(d{y)}(({z}(y,x))(e^{{x}{z}(y,x)}+{x}{y})) is {z}(y,x)(0+{x})
What is the first d/(d{y)}(({z}(y,x))(e^{{x}{z}(y,x)}+{x}{y})) ?
The first d/(d{y)}(({z}(y,x))(e^{{x}{z}(y,x)}+{x}{y})) is {z}(y,x)(0+{x})