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

The derivative of f(x)=(2x^2-5x)/(3x+5) is (6x^2+20x-25)/((3x+5)^2)

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

The first derivative of f(x)=(2x^2-5x)/(3x+5) is (6x^2+20x-25)/((3x+5)^2)

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derivative of

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

\frac{6 x ^{2} + 20 x - 25}{( 3 x + 5 ) ^{2}}

Solution steps

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

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

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

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

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

Expand

(4 x - 5) (3 x + 5) - 3 (2 x^{2} - 5 x) : 6 x^{2} + 20 x - 25

= \frac{6 x ^{2} + 20 x - 25}{( 3 x + 5 ) ^{2}}

\int_{- 5}^{5} x d x

\frac{d y}{d x} = \frac{4 y ^{2} - x ^{4}}{4 x y}

tan (7 x)

\int 4 (x - 1) ln (x - 1) d x

y^{^{'}} - \frac{3}{x} y = \frac{y ^{4}}{x ^{7}}

What is the derivative of f(x)=(2x^2-5x)/(3x+5) ?
The derivative of f(x)=(2x^2-5x)/(3x+5) is (6x^2+20x-25)/((3x+5)^2)
What is the first derivative of f(x)=(2x^2-5x)/(3x+5) ?
The first derivative of f(x)=(2x^2-5x)/(3x+5) is (6x^2+20x-25)/((3x+5)^2)