What is (\partial)/(\partial x)(e^xy+5xy^2-x^2) ?

The answer to (\partial)/(\partial x)(e^xy+5xy^2-x^2) is ye^x+5y^2-2x

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(\partial)/(\partial x)(e^xy+5xy^2-x^2)

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

y e^{x} + 5 y^{2} - 2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (e^{x} y) = y e^{x}

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

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

= y e^{x} + 5 y^{2} - 2 x

n = 0 \sum \infty \frac{3}{n ^{4}}

\frac{d}{d x} (x cos (x) - sin (x))

\int sin (t) cos (n t) d t

y^{^{''}} - 3 y^{^{'}} = e^{3 t}

k = 1 \sum \infty \frac{5 + ( - 2 ) ^{k}}{4 ^{k + 1}}

What is (\partial)/(\partial x)(e^xy+5xy^2-x^2) ?
The answer to (\partial)/(\partial x)(e^xy+5xy^2-x^2) is ye^x+5y^2-2x