What is the derivative of 3e^{-y}(x^2+y^2)+4 ?

The derivative of 3e^{-y}(x^2+y^2)+4 is 3(2e^{-y}y-e^{-y}(x^2+y^2))

What is the first derivative of 3e^{-y}(x^2+y^2)+4 ?

The first derivative of 3e^{-y}(x^2+y^2)+4 is 3(2e^{-y}y-e^{-y}(x^2+y^2))

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derivative of

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

3 (2 e^{- y} y - e^{- y} (x^{2} + y^{2}))

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 3 (- e^{- y} (x^{2} + y^{2}) + 2 e^{- y} y) + 0

Simplify

= 3 (2 e^{- y} y - e^{- y} (x^{2} + y^{2}))

\int (3 x^{2} - 5 x^{3} + 4) d x

\int (x^{3}) d x

\int \frac{3}{x ^{2} - 2 x - 3} d x

\int \frac{1}{x ^{\frac{1}{3}} + x ^{\frac{1}{2}}} d x

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

What is the derivative of 3e^{-y}(x^2+y^2)+4 ?
The derivative of 3e^{-y}(x^2+y^2)+4 is 3(2e^{-y}y-e^{-y}(x^2+y^2))
What is the first derivative of 3e^{-y}(x^2+y^2)+4 ?
The first derivative of 3e^{-y}(x^2+y^2)+4 is 3(2e^{-y}y-e^{-y}(x^2+y^2))