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

The derivative of (3x^2/(2x+1)) is (3(2x^2+2x))/((2x+1)^2)

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

The first derivative of (3x^2/(2x+1)) is (3(2x^2+2x))/((2x+1)^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{2}) = 2 x

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

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

Simplify

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

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

\frac{d y}{d x} = \frac{( x - 2 )}{( y - 2 )}

(1 + cos (t)) \frac{d y}{d t} = (1 - e^{- y}) sin (t)

y = lo g_{5} (x) + lo g_{5} (x^{4})

n = 1 \sum \infty e^{- n} n^{3}

(2 x y^{2} - 4) d x + (2 x^{2} y + 7) d y = 0

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