What is (\partial)/(\partial x)((-x-3)/(y^2)) ?

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

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(\partial)/(\partial x)((-x-3)/(y^2))

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

- \frac{1}{y ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (3) = 0

= \frac{1}{y ^{2}} (- 1 - 0)

Simplify

= - \frac{1}{y ^{2}}

(2 x + 1)^{^{'}}

\frac{d}{d x} (2 x^{4} + x^{3} + 3 x^{2})

f (x) = \frac{1}{x - 3}, at x = 8

\int t^{2} sin (t) d t

f (t) = cos^{2} (\frac{π}{2} - t)

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