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

The derivative of (3x^2-1/(2x+5)) is (6x^2+30x+2)/((2x+5)^2)

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

The first derivative of (3x^2-1/(2x+5)) is (6x^2+30x+2)/((2x+5)^2)

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

\frac{6 x ^{2} + 30 x + 2}{( 2 x + 5 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (3 x^{2} - 1) = 6 x

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

= \frac{6 x ( 2 x + 5 ) - 2 ( 3 x ^{2} - 1 )}{( 2 x + 5 ) ^{2}}

Expand

6 x (2 x + 5) - 2 (3 x^{2} - 1) : 6 x^{2} + 30 x + 2

= \frac{6 x ^{2} + 30 x + 2}{( 2 x + 5 ) ^{2}}

n = 0 \sum \infty 6 (\frac{4}{5})^{n}

\int \frac{tan ( ln ( 4 x + 2 ) )}{4 x + 2} d x

(5 x + 3) x^{2}

n = 0 \sum \infty (\frac{2}{3}) (\frac{5}{4})^{n}

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

What is the derivative of (3x^2-1/(2x+5)) ?
The derivative of (3x^2-1/(2x+5)) is (6x^2+30x+2)/((2x+5)^2)
What is the first derivative of (3x^2-1/(2x+5)) ?
The first derivative of (3x^2-1/(2x+5)) is (6x^2+30x+2)/((2x+5)^2)