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

The derivative of (3x^2-1/(6x^3+5)) is (-18x^4+18x^2+30x)/((6x^3+5)^2)

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

The first derivative of (3x^2-1/(6x^3+5)) is (-18x^4+18x^2+30x)/((6x^3+5)^2)

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

\frac{- 18 x ^{4} + 18 x ^{2} + 30 x}{( 6 x ^{3} + 5 ) ^{2}}

Solution steps

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

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

\frac{d}{d x} (6 x^{3} + 5) = 18 x^{2}

= \frac{6 x ( 6 x ^{3} + 5 ) - 18 x ^{2} ( 3 x ^{2} - 1 )}{( 6 x ^{3} + 5 ) ^{2}}

Expand

6 x (6 x^{3} + 5) - 18 x^{2} (3 x^{2} - 1) : - 18 x^{4} + 18 x^{2} + 30 x

= \frac{- 18 x ^{4} + 18 x ^{2} + 30 x}{( 6 x ^{3} + 5 ) ^{2}}

x \to 7 + lim (\frac{3}{x - 7})

\int_{5}^{\infty} \frac{1}{x ^{2} - 4} d x

\frac{d}{d x} (\frac{1}{4 x ^{2} + 2 x})

x \to 2 lim (2 + 3)

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

What is the derivative of (3x^2-1/(6x^3+5)) ?
The derivative of (3x^2-1/(6x^3+5)) is (-18x^4+18x^2+30x)/((6x^3+5)^2)
What is the first derivative of (3x^2-1/(6x^3+5)) ?
The first derivative of (3x^2-1/(6x^3+5)) is (-18x^4+18x^2+30x)/((6x^3+5)^2)