What is the derivative of xe^y+y^2e^x-1/3 x+4 ?

The derivative of xe^y+y^2e^x-1/3 x+4 is e^yx(dy)/(dx)+e^y+e^xy^2+2e^xy(dy)/(dx)-1/3

What is the first derivative of xe^y+y^2e^x-1/3 x+4 ?

The first derivative of xe^y+y^2e^x-1/3 x+4 is e^yx(dy)/(dx)+e^y+e^xy^2+2e^xy(dy)/(dx)-1/3

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\frac{d}{d x} (x e^{y} + y^{2} e^{x} - \frac{1}{3} x + 4)

e^{y} x \frac{d y}{d x} + e^{y} + e^{x} y^{2} + 2 e^{x} y \frac{d y}{d x} - \frac{1}{3}

Solution steps

\frac{d}{d x} (x e^{y} + y^{2} e^{x} - \frac{1}{3} x + 4)

Simplify

x e^{y} + y^{2} e^{x} - \frac{1}{3} x + 4 : e^{y} x + e^{x} y^{2} - \frac{x}{3} + 4

= \frac{d}{d x} (e^{y} x + e^{x} y^{2} - \frac{x}{3} + 4)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (e^{y} x) + \frac{d}{d x} (e^{x} y^{2}) - \frac{d}{d x} (\frac{x}{3}) + \frac{d}{d x} (4)

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

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

\frac{d}{d x} (\frac{x}{3}) = \frac{1}{3}

\frac{d}{d x} (4) = 0

= e^{y} x \frac{d y}{d x} + e^{y} + e^{x} y^{2} + 2 y \frac{d y}{d x} e^{x} - \frac{1}{3} + 0

Simplify

= e^{y} x \frac{d y}{d x} + e^{y} + e^{x} y^{2} + 2 e^{x} y \frac{d y}{d x} - \frac{1}{3}

\int x e^{3 x^{2}} d x

\frac{d}{d x} (e^{i x})

\int \frac{2 cos ( ln ( x ) )}{x} d x

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

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

What is the derivative of xe^y+y^2e^x-1/3 x+4 ?
The derivative of xe^y+y^2e^x-1/3 x+4 is e^yx(dy)/(dx)+e^y+e^xy^2+2e^xy(dy)/(dx)-1/3
What is the first derivative of xe^y+y^2e^x-1/3 x+4 ?
The first derivative of xe^y+y^2e^x-1/3 x+4 is e^yx(dy)/(dx)+e^y+e^xy^2+2e^xy(dy)/(dx)-1/3