What is (\partial)/(\partial v)(u/(u+v)) ?

The answer to (\partial)/(\partial v)(u/(u+v)) is -u/((u+v)^2)

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(\partial)/(\partial v)(u/(u+v))

\frac{\partial}{\partial v} (\frac{u}{u + v})

- \frac{u}{( u + v ) ^{2}}

Solution steps

\frac{\partial}{\partial v} (\frac{u}{u + v})

Treat

u

as a constant

Take the constant out:

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

= u \frac{\partial}{\partial v} (\frac{1}{u + v})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= u \frac{\partial}{\partial v} ((u + v)^{- 1})

Apply the chain rule:

- \frac{1}{( u + v ) ^{2}} \frac{\partial}{\partial v} (u + v)

= - \frac{1}{( u + v ) ^{2}} \frac{\partial}{\partial v} (u + v)

\frac{\partial}{\partial v} (u + v) = 1

= u (- \frac{1}{( u + v ) ^{2}} \cdot 1)

Simplify

u (- \frac{1}{( u + v ) ^{2}} \cdot 1) : - \frac{u}{( u + v ) ^{2}}

= - \frac{u}{( u + v ) ^{2}}

y^{^{'}} = e^{6 x} - 4 x

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

\frac{\partial}{\partial x} (x^{2} - 4 x y + y^{3} + 4 y)

\frac{- 200 ln ( \frac{f}{40} )}{( f + 20 )}

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

What is (\partial)/(\partial v)(u/(u+v)) ?
The answer to (\partial)/(\partial v)(u/(u+v)) is -u/((u+v)^2)