What is the derivative of f(x)=((2x+1)/(x+5))(3x-1) ?

The derivative of f(x)=((2x+1)/(x+5))(3x-1) is (6x^2+60x+6)/((x+5)^2)

What is the first derivative of f(x)=((2x+1)/(x+5))(3x-1) ?

The first derivative of f(x)=((2x+1)/(x+5))(3x-1) is (6x^2+60x+6)/((x+5)^2)

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derivative of

f (x) = (\frac{2 x + 1}{x + 5}) (3 x - 1)

\frac{6 x ^{2} + 60 x + 6}{( x + 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x + 1}{x + 5} (3 x - 1))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{2 x + 1}{x + 5}, g = 3 x - 1

= \frac{d}{d x} (\frac{2 x + 1}{x + 5}) (3 x - 1) + \frac{d}{d x} (3 x - 1) \frac{2 x + 1}{x + 5}

\frac{d}{d x} (\frac{2 x + 1}{x + 5}) = \frac{9}{( x + 5 ) ^{2}}

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

= \frac{9}{( x + 5 ) ^{2}} (3 x - 1) + 3 \cdot \frac{2 x + 1}{x + 5}

Simplify

\frac{9}{( x + 5 ) ^{2}} (3 x - 1) + 3 \cdot \frac{2 x + 1}{x + 5} : \frac{6 x ^{2} + 60 x + 6}{( x + 5 ) ^{2}}

= \frac{6 x ^{2} + 60 x + 6}{( x + 5 ) ^{2}}

\frac{d ^{2}}{d x ^{2}} (\frac{6}{7 t ^{4}})

x^{2} ln (2 x)

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

\int (3 x^{3} + 4 x^{2} - 2 x^{- 1} + 13) d x

t^{2} cos (5 t)

What is the derivative of f(x)=((2x+1)/(x+5))(3x-1) ?
The derivative of f(x)=((2x+1)/(x+5))(3x-1) is (6x^2+60x+6)/((x+5)^2)
What is the first derivative of f(x)=((2x+1)/(x+5))(3x-1) ?
The first derivative of f(x)=((2x+1)/(x+5))(3x-1) is (6x^2+60x+6)/((x+5)^2)