What is the derivative of (4x^3-5x^2/(3x^2-4)) ?

The derivative of (4x^3-5x^2/(3x^2-4)) is (12x^4-48x^2+40x)/((3x^2-4)^2)

What is the first derivative of (4x^3-5x^2/(3x^2-4)) ?

The first derivative of (4x^3-5x^2/(3x^2-4)) is (12x^4-48x^2+40x)/((3x^2-4)^2)

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

\frac{12 x ^{4} - 48 x ^{2} + 40 x}{( 3 x ^{2} - 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{3} - 5 x^{2}) = 12 x^{2} - 10 x

\frac{d}{d x} (3 x^{2} - 4) = 6 x

= \frac{( 12 x ^{2} - 10 x ) ( 3 x ^{2} - 4 ) - 6 x ( 4 x ^{3} - 5 x ^{2} )}{( 3 x ^{2} - 4 ) ^{2}}

Expand

(12 x^{2} - 10 x) (3 x^{2} - 4) - 6 x (4 x^{3} - 5 x^{2}) : 12 x^{4} - 48 x^{2} + 40 x

= \frac{12 x ^{4} - 48 x ^{2} + 40 x}{( 3 x ^{2} - 4 ) ^{2}}

\int_{- 1}^{1} (20 - 12 x)^{- 1} d x

\int \frac{( sin ( 2 x ) )}{e ^{x}} d x

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

x (4 - x)^{3}

\frac{d}{d t} (e^{t^{3}})

What is the derivative of (4x^3-5x^2/(3x^2-4)) ?
The derivative of (4x^3-5x^2/(3x^2-4)) is (12x^4-48x^2+40x)/((3x^2-4)^2)
What is the first derivative of (4x^3-5x^2/(3x^2-4)) ?
The first derivative of (4x^3-5x^2/(3x^2-4)) is (12x^4-48x^2+40x)/((3x^2-4)^2)