What is the d/(dy)(e^{x^2-y^2}) ?

The d/(dy)(e^{x^2-y^2}) is -2e^{x^2-y^2}y

What is the first d/(dy)(e^{x^2-y^2}) ?

The first d/(dy)(e^{x^2-y^2}) is -2e^{x^2-y^2}y

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

- 2 e^{x^{2} - y^{2}} y

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

= - 2 e^{x^{2} - y^{2}} y

\int - 7 x^{3} (6 x^{4} + 1)^{6} d x

x \to - \infty lim (\frac{1}{4 x})

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

n = 1 \sum \infty \frac{x ^{n}}{n + 1}

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

What is the d/(dy)(e^{x^2-y^2}) ?
The d/(dy)(e^{x^2-y^2}) is -2e^{x^2-y^2}y
What is the first d/(dy)(e^{x^2-y^2}) ?
The first d/(dy)(e^{x^2-y^2}) is -2e^{x^2-y^2}y