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

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

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

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

z e^{x^{2} - y^{2}} \cdot 2 x

Solution steps

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

Treat

z, y

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

= z e^{x^{2} - y^{2}} \cdot 2 x

y^{^{''}} - 4 y^{^{'}} - 4 y = e^{2 x}

y = \frac{x ^{2}}{2} - 4 x + 6, y = \frac{x}{2} - 1, x = 0, x = 5

\int \frac{441}{x ^{3} - 21 x ^{2}} d x

\int \frac{x ^{2}}{( 25 + x ^{2} ) ^{2}} d x

5 x y \frac{d y}{d x} = 1

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