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})

Take the constant out:

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

= - \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}}

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

y^{^{'}} - 2 y + x^{2} = 0

\int 32 x d x

\frac{d}{d x} (\frac{1}{5 x ^{2} + 4 x})

\int (\frac{y}{3 y + 5}) d y

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)