What is the derivative of f(x)=((2+3x)/(3x))(9-4x) ?

The derivative of f(x)=((2+3x)/(3x))(9-4x) is -(2(2x^2+3))/(x^2)

What is the first derivative of f(x)=((2+3x)/(3x))(9-4x) ?

The first derivative of f(x)=((2+3x)/(3x))(9-4x) is -(2(2x^2+3))/(x^2)

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derivative of

f (x) = (\frac{2 + 3 x}{3 x}) (9 - 4 x)

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

Solution steps

\frac{d}{d x} (\frac{2 + 3 x}{3 x} (9 - 4 x))

Take the constant out:

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

= \frac{1}{3} \frac{d}{d x} (\frac{2 + 3 x}{x} (9 - 4 x))

Apply the Product Rule:

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

f = \frac{2 + 3 x}{x}, g = 9 - 4 x

= \frac{1}{3} (\frac{d}{d x} (\frac{2 + 3 x}{x}) (9 - 4 x) + \frac{d}{d x} (9 - 4 x) \frac{2 + 3 x}{x})

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

\frac{d}{d x} (9 - 4 x) = - 4

= \frac{1}{3} ((- \frac{2}{x ^{2}}) (9 - 4 x) + (- 4) \frac{2 + 3 x}{x})

Simplify

\frac{1}{3} ((- \frac{2}{x ^{2}}) (9 - 4 x) + (- 4) \frac{2 + 3 x}{x}) : - \frac{2 ( 2 x ^{2} + 3 )}{x ^{2}}

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

\frac{d}{d x} (5 sin (x) cos (x))

\int (x + 1) (9 x - 8) d x

\int \frac{x ^{7} - 4 x}{x ^{5}} d x

sin (10 x)

\frac{d}{d x} (e^{1 + s i n (2 x)})

What is the derivative of f(x)=((2+3x)/(3x))(9-4x) ?
The derivative of f(x)=((2+3x)/(3x))(9-4x) is -(2(2x^2+3))/(x^2)
What is the first derivative of f(x)=((2+3x)/(3x))(9-4x) ?
The first derivative of f(x)=((2+3x)/(3x))(9-4x) is -(2(2x^2+3))/(x^2)