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

The derivative of f(x)=(2x^2-5x+3)/(5x-3) is (10x^2-12x)/((5x-3)^2)

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

The first derivative of f(x)=(2x^2-5x+3)/(5x-3) is (10x^2-12x)/((5x-3)^2)

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derivative of

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

\frac{10 x ^{2} - 12 x}{( 5 x - 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(4 x - 5) (5 x - 3) - 5 (2 x^{2} - 5 x + 3) : 10 x^{2} - 12 x

= \frac{10 x ^{2} - 12 x}{( 5 x - 3 ) ^{2}}

\int \frac{e ^{x} + 1}{x} d x

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

\int \frac{65}{( x - 1 ) ( x ^{2} + 64 )} d x

x \to 25 lim (x - 25)

(x + 1) d y = (x + 6) d x

What is the derivative of f(x)=(2x^2-5x+3)/(5x-3) ?
The derivative of f(x)=(2x^2-5x+3)/(5x-3) is (10x^2-12x)/((5x-3)^2)
What is the first derivative of f(x)=(2x^2-5x+3)/(5x-3) ?
The first derivative of f(x)=(2x^2-5x+3)/(5x-3) is (10x^2-12x)/((5x-3)^2)