What is the derivative of 1/(x+c) ?

The derivative of 1/(x+c) is -1/((x+c)^2)

What is the first derivative of 1/(x+c) ?

The first derivative of 1/(x+c) is -1/((x+c)^2)

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int (2 x + 1)^{9} d x

\frac{d}{d t} (t^{2} + 2)

\int 3 x e^{3 x} d x

f (x) = \frac{cos ( t )}{t ^{7}}

x^{^{''}} - 3 x^{^{'}} = e^{- 3 t}

What is the derivative of 1/(x+c) ?
The derivative of 1/(x+c) is -1/((x+c)^2)
What is the first derivative of 1/(x+c) ?
The first derivative of 1/(x+c) is -1/((x+c)^2)