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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

\frac{d}{d x} (1.5 (\frac{1}{3} + \frac{x}{2} - \frac{x ^{3}}{6}))

\int \frac{1}{- x ^{2} - 2 x 8} d x

\int \frac{( 1 )}{x ^{4}} d x

(x^{2})^{^{'}}

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

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