What is the derivative of f(x)= 4/(x+1) ?

The derivative of f(x)= 4/(x+1) is -4/((x+1)^2)

What is the first derivative of f(x)= 4/(x+1) ?

The first derivative of f(x)= 4/(x+1) is -4/((x+1)^2)

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derivative of

f (x) = \frac{4}{x + 1}

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

Solution steps

\frac{d}{d x} (\frac{4}{x + 1})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{1}{x + 1})

Apply exponent rule:

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

= 4 \frac{d}{d x} ((x + 1)^{- 1})

Apply the chain rule:

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

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

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

= 4 (- \frac{1}{( x + 1 ) ^{2}} \cdot 1)

Simplify

4 (- \frac{1}{( x + 1 ) ^{2}} \cdot 1) : - \frac{4}{( x + 1 ) ^{2}}

= - \frac{4}{( x + 1 ) ^{2}}

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

f (x) = x + \frac{4}{x}, at x = 4

\frac{d}{d z} (ln (\frac{c ^{2} - z ^{2}}{c ^{2} + z ^{2}}))

y^{^{''}} + 4 y^{^{'}} + 6 y = 1 + e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} ((4 x + 7 y)^{10})

What is the derivative of f(x)= 4/(x+1) ?
The derivative of f(x)= 4/(x+1) is -4/((x+1)^2)
What is the first derivative of f(x)= 4/(x+1) ?
The first derivative of f(x)= 4/(x+1) is -4/((x+1)^2)