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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

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