What is (\partial)/(\partial x)((e^{7-x})/(y^5)) ?

The answer to (\partial)/(\partial x)((e^{7-x})/(y^5)) is -(e^{7-x})/(y^5)

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(\partial)/(\partial x)((e^{7-x})/(y^5))

\frac{\partial}{\partial x} (\frac{e ^{7 - x}}{y ^{5}})

- \frac{e ^{7 - x}}{y ^{5}}

Solution steps

\frac{\partial}{\partial x} (\frac{e ^{7 - x}}{y ^{5}})

Treat

y

as a constant

Take the constant out:

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

= \frac{1}{y ^{5}} \frac{\partial}{\partial x} (e^{7 - x})

Apply the chain rule:

e^{7 - x} \frac{\partial}{\partial x} (7 - x)

= e^{7 - x} \frac{\partial}{\partial x} (7 - x)

\frac{\partial}{\partial x} (7 - x) = - 1

= \frac{1}{y ^{5}} e^{7 - x} (- 1)

Simplify

\frac{1}{y ^{5}} e^{7 - x} (- 1) : - \frac{e ^{7 - x}}{y ^{5}}

= - \frac{e ^{7 - x}}{y ^{5}}

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

n = 0 \sum \infty \frac{1}{n ^{\frac{2}{3}}}

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

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

\frac{x}{8}

What is (\partial)/(\partial x)((e^{7-x})/(y^5)) ?
The answer to (\partial)/(\partial x)((e^{7-x})/(y^5)) is -(e^{7-x})/(y^5)