What is (\partial)/(\partial x)((c+1)/(c+x)) ?

The answer to (\partial)/(\partial x)((c+1)/(c+x)) is -(c+1)/((c+x)^2)

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(\partial)/(\partial x)((c+1)/(c+x))

\frac{\partial}{\partial x} (\frac{c + 1}{c + x})

- \frac{c + 1}{( c + x ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{c + 1}{c + x})

Take the constant out:

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

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

Apply exponent rule:

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

= (c + 1) \frac{\partial}{\partial x} ((c + x)^{- 1})

Apply the chain rule:

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

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

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

= (c + 1) (- \frac{1}{( c + x ) ^{2}} \cdot 1)

Simplify

(c + 1) (- \frac{1}{( c + x ) ^{2}} \cdot 1) : - \frac{c + 1}{( c + x ) ^{2}}

= - \frac{c + 1}{( c + x ) ^{2}}

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

\frac{d y}{d x} = 3 x y

f (x) = arccos (x^{2})

(0, - 2), (1, 2)

\frac{\partial}{\partial u} (\frac{1}{3} (v - 2 u))

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