What is (\partial)/(\partial x)(3e^{-y}(x^2+y^2)+4) ?

The answer to (\partial)/(\partial x)(3e^{-y}(x^2+y^2)+4) is 6e^{-y}x

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(\partial)/(\partial x)(3e^{-y}(x^2+y^2)+4)

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

6 e^{- y} x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 e^{- y} (x^{2} + y^{2})) = 6 e^{- y} x

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

= 6 e^{- y} x + 0

Simplify

= 6 e^{- y} x

\int \frac{1}{x ( x ^{2} - 16 ) ^{\frac{3}{2}}} d x

f (x) = \frac{cos ( x )}{1 + a sin ( x )}

y = sin (ln (x))

\int_{0}^{1} x^{2} d x

\frac{\partial}{\partial x} (x e^{x y} cos (2 x) - 3)

What is (\partial)/(\partial x)(3e^{-y}(x^2+y^2)+4) ?
The answer to (\partial)/(\partial x)(3e^{-y}(x^2+y^2)+4) is 6e^{-y}x