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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

x \to 1 lim (\frac{x ^{3} - x}{1 - x})

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

6 cos (sin (x^{2}))

\frac{d}{d x} (ln (1 + 0))

\int \frac{x}{4 - x} d x

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