What is (\partial)/(\partial p)(((p+q))/(p+r)) ?

The answer to (\partial)/(\partial p)(((p+q))/(p+r)) is (-q+r)/((p+r)^2)

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(\partial)/(\partial p)(((p+q))/(p+r))

\frac{\partial}{\partial p} (\frac{( p + q )}{p + r})

\frac{- q + r}{( p + r ) ^{2}}

Solution steps

\frac{\partial}{\partial p} (\frac{( p + q )}{p + r})

Treat

q, r

as constants

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{\partial}{\partial p} ( p + q ) ( p + r ) - \frac{\partial}{\partial p} ( p + r ) ( p + q )}{( p + r ) ^{2}}

\frac{\partial}{\partial p} (p + q) = 1

\frac{\partial}{\partial p} (p + r) = 1

= \frac{1 \cdot ( p + r ) - 1 \cdot ( p + q )}{( p + r ) ^{2}}

Simplify

\frac{1 \cdot ( p + r ) - 1 \cdot ( p + q )}{( p + r ) ^{2}} : \frac{- q + r}{( p + r ) ^{2}}

= \frac{- q + r}{( p + r ) ^{2}}

x \to 0 lim (\frac{sin ( 4 x )}{cos ( 5 x )})

\frac{d}{d x} (- 3 x^{3} + 2 x^{2} + 5 x - 2)

x \to 0 lim ((x^{2}) sin (\frac{1}{x}))

\int \frac{x ^{2} + x + 1}{x ^{3} + 2 x ^{2}} d x

\frac{\partial}{\partial y} (x e^{x} cos (y))

What is (\partial)/(\partial p)(((p+q))/(p+r)) ?
The answer to (\partial)/(\partial p)(((p+q))/(p+r)) is (-q+r)/((p+r)^2)