What is (\partial)/(\partial y)(e^{1+x^2-y^2}) ?

The answer to (\partial)/(\partial y)(e^{1+x^2-y^2}) is -2e^{1+x^2-y^2}y

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(\partial)/(\partial y)(e^{1+x^2-y^2})

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

- 2 e^{1 + x^{2} - y^{2}} y

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

= - 2 e^{1 + x^{2} - y^{2}} y

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

\int_{0}^{2} 6 x^{2} - 18 x + 12 d x

f (x) = x^{2} - \frac{1}{x}, (2, 3.5)

n = 1 \sum \infty \frac{( ( - 7 ) ^{n - 1} )}{8 ^{n}}

y^{^{''}} - 14 y^{^{'}} + 49 y = t^{- 2} e^{7 t}

What is (\partial)/(\partial y)(e^{1+x^2-y^2}) ?
The answer to (\partial)/(\partial y)(e^{1+x^2-y^2}) is -2e^{1+x^2-y^2}y