What is the derivative of f(x)=(6x)/(x+1) ?

The derivative of f(x)=(6x)/(x+1) is 6/((x+1)^2)

What is the first derivative of f(x)=(6x)/(x+1) ?

The first derivative of f(x)=(6x)/(x+1) is 6/((x+1)^2)

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derivative of

f (x) = \frac{6 x}{x + 1}

\frac{6}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{x + 1})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{x + 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 6 \cdot \frac{\frac{d x}{d x} ( x + 1 ) - \frac{d}{d x} ( x + 1 ) x}{( x + 1 ) ^{2}}

\frac{d x}{d x} = 1

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

= 6 \cdot \frac{1 \cdot ( x + 1 ) - 1 \cdot x}{( x + 1 ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( x + 1 ) - 1 \cdot x}{( x + 1 ) ^{2}} : \frac{6}{( x + 1 ) ^{2}}

= \frac{6}{( x + 1 ) ^{2}}

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

\int \frac{x ^{5}}{x ^{6} + 5} d x

(e^{4 t} - e^{- 3 t})^{2}

\frac{d}{d x} ((x^{2} - 4)^{5} (3 x + 5)^{4})

\int \frac{5 x - 3}{( x + 1 ) ( x - 3 )} d x

What is the derivative of f(x)=(6x)/(x+1) ?
The derivative of f(x)=(6x)/(x+1) is 6/((x+1)^2)
What is the first derivative of f(x)=(6x)/(x+1) ?
The first derivative of f(x)=(6x)/(x+1) is 6/((x+1)^2)