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

The answer to (\partial)/(\partial x)(-1/(x^2)+e^{-y}) is 2/(x^3)

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

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

\frac{2}{x ^{3}}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial x} (\frac{1}{x ^{2}}) = - \frac{2}{x ^{3}}

\frac{\partial}{\partial x} (e^{- y}) = 0

= - (- \frac{2}{x ^{3}}) + 0

Simplify

= \frac{2}{x ^{3}}

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

\int \frac{1}{( x + 3 ) ( x + 5 )} d x

\frac{\partial}{\partial x} ((5 x - x^{2}) cos (t))

Y (s) = \frac{s - 2}{s ^{2} - 2 s - 3}

f (x) = - 3 x^{2} + x + 2, at x = 1

What is (\partial)/(\partial x)(-1/(x^2)+e^{-y}) ?
The answer to (\partial)/(\partial x)(-1/(x^2)+e^{-y}) is 2/(x^3)