What is the derivative of f(x)=((x+1)^2)/(3x-4) ?

The derivative of f(x)=((x+1)^2)/(3x-4) is (3x^2-8x-11)/((3x-4)^2)

What is the first derivative of f(x)=((x+1)^2)/(3x-4) ?

The first derivative of f(x)=((x+1)^2)/(3x-4) is (3x^2-8x-11)/((3x-4)^2)

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derivative of

f (x) = \frac{( x + 1 ) ^{2}}{3 x - 4}

\frac{3 x ^{2} - 8 x - 11}{( 3 x - 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( x + 1 ) ^{2}}{3 x - 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ( x + 1 ) ^{2} ) ( 3 x - 4 ) - \frac{d}{d x} ( 3 x - 4 ) ( x + 1 ) ^{2}}{( 3 x - 4 ) ^{2}}

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

\frac{d}{d x} (3 x - 4) = 3

= \frac{2 ( x + 1 ) ( 3 x - 4 ) - 3 ( x + 1 ) ^{2}}{( 3 x - 4 ) ^{2}}

Expand

2 (x + 1) (3 x - 4) - 3 (x + 1)^{2} : 3 x^{2} - 8 x - 11

= \frac{3 x ^{2} - 8 x - 11}{( 3 x - 4 ) ^{2}}

\int \frac{3 - 18 x}{4 - 9 x ^{2}} d x

\int \frac{1}{( 1 - 6 x ) ^{4}} d x

\frac{( s + 1 )}{( s ^{2} + s + 1 )}

x \to + 4 lim (5)

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

What is the derivative of f(x)=((x+1)^2)/(3x-4) ?
The derivative of f(x)=((x+1)^2)/(3x-4) is (3x^2-8x-11)/((3x-4)^2)
What is the first derivative of f(x)=((x+1)^2)/(3x-4) ?
The first derivative of f(x)=((x+1)^2)/(3x-4) is (3x^2-8x-11)/((3x-4)^2)