What is the derivative of 1/(3+x) ?

The derivative of 1/(3+x) is -1/((3+x)^2)

What is the first derivative of 1/(3+x) ?

The first derivative of 1/(3+x) is -1/((3+x)^2)

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derivative of

\frac{1}{3 + x}

- \frac{1}{( 3 + x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{3 + x})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 + x) = 1

= - \frac{1}{( 3 + x ) ^{2}} \cdot 1

Simplify

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

\frac{x}{x ^{2} - 1}

\int 6 e^{3 x} - \frac{8}{e ^{2 x}} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = (x + 1) e^{2 x}

f (x) = \frac{1}{8 x}, at (x = 9)

n = 0 \sum \infty \frac{2 ^{n}}{( 2 n ) !} x^{2 n}

What is the derivative of 1/(3+x) ?
The derivative of 1/(3+x) is -1/((3+x)^2)
What is the first derivative of 1/(3+x) ?
The first derivative of 1/(3+x) is -1/((3+x)^2)