What is (\partial)/(\partial x)(2x^y+8y^x+10) ?

The answer to (\partial)/(\partial x)(2x^y+8y^x+10) is 2yx^{y-1}+8y^xln(y)

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(\partial)/(\partial x)(2x^y+8y^x+10)

\frac{\partial}{\partial x} (2 x^{y} + 8 y^{x} + 10)

2 y x^{y - 1} + 8 y^{x} ln (y)

Solution steps

\frac{\partial}{\partial x} (2 x^{y} + 8 y^{x} + 10)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (2 x^{y}) + \frac{\partial}{\partial x} (8 y^{x}) + \frac{\partial}{\partial x} (10)

\frac{\partial}{\partial x} (2 x^{y}) = 2 y x^{y - 1}

\frac{\partial}{\partial x} (8 y^{x}) = 8 y^{x} ln (y)

\frac{\partial}{\partial x} (10) = 0

= 2 y x^{y - 1} + 8 y^{x} ln (y) + 0

Simplify

= 2 y x^{y - 1} + 8 y^{x} ln (y)

x \to - 2 - lim (- 1 - x + 1)

y^{9} y^{^{'}} = 11 cos (π x)

\int_{- \frac{w}{2}}^{0} (1 + \frac{2}{w} x) e^{- ik x} d x

y^{^{''}} + 5 y^{^{'}} - 36 y = 0

\frac{d ^{2} y}{d x ^{2}} + 14 \frac{d y}{d x} + 49 y = 0

What is (\partial)/(\partial x)(2x^y+8y^x+10) ?
The answer to (\partial)/(\partial x)(2x^y+8y^x+10) is 2yx^{y-1}+8y^xln(y)