What is (\partial)/(\partial x)(2xe^{x^2}+y^2) ?

The answer to (\partial)/(\partial x)(2xe^{x^2}+y^2) is 2(2e^{x^2}x^2+e^{x^2})

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(\partial)/(\partial x)(2xe^{x^2}+y^2)

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

2 (2 e^{x^{2}} x^{2} + e^{x^{2}})

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (y^{2}) = 0

= 2 (e^{x^{2}} + 2 e^{x^{2}} x^{2}) + 0

Simplify

= 2 (2 e^{x^{2}} x^{2} + e^{x^{2}})

y^{^{''}} + 6 y^{^{'}} + 18 y = 0

y \to 0 lim (cot (5 y))

\frac{\partial}{\partial x} (e^{6 y} cos (- 6 x))

\frac{d y}{d x} = (\frac{1 + 2 y ^{2}}{y sin ( x )})

\frac{d y}{d x} = \frac{x ( y - 2 )}{x ^{2} + 4}, y (1) = 5

What is (\partial)/(\partial x)(2xe^{x^2}+y^2) ?
The answer to (\partial)/(\partial x)(2xe^{x^2}+y^2) is 2(2e^{x^2}x^2+e^{x^2})