What is the derivative of (2x^2/(6x+1)) ?

The derivative of (2x^2/(6x+1)) is (2(6x^2+2x))/((6x+1)^2)

What is the first derivative of (2x^2/(6x+1)) ?

The first derivative of (2x^2/(6x+1)) is (2(6x^2+2x))/((6x+1)^2)

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\frac{d}{d x} (\frac{2 x ^{2}}{6 x + 1})

\frac{2 ( 6 x ^{2} + 2 x )}{( 6 x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x ^{2}}{6 x + 1})

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (6 x + 1) = 6

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

Simplify

2 \cdot \frac{2 x ( 6 x + 1 ) - 6 x ^{2}}{( 6 x + 1 ) ^{2}} : \frac{2 ( 6 x ^{2} + 2 x )}{( 6 x + 1 ) ^{2}}

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

x \to \infty lim (\frac{x ^{2} + 2 ^{x}}{( x + 1 ) ^{2} + 2 ^{x + 1}})

\frac{d}{d t} (\frac{3 e ^{3 α t} xα}{4})

\int (x + 5) 7 d x

\int \frac{3}{1 - y} d y

y = - x^{2} + 1, (- 1, 1)

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