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

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

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

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

- e^{1 + y - x}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

= - e^{1 + y - x}

\int 3 x - \frac{2}{x} d x

y^{^{'}} = y - t, y (0) = 0.5

\int (\frac{4}{x ^{3}}) d x

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

f (x) = (x - 2)^{3} + 10

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