What is the derivative of (5x^4+3x^3/(4x-9)) ?

The derivative of (5x^4+3x^3/(4x-9)) is (60x^4-156x^3-81x^2)/((4x-9)^2)

What is the first derivative of (5x^4+3x^3/(4x-9)) ?

The first derivative of (5x^4+3x^3/(4x-9)) is (60x^4-156x^3-81x^2)/((4x-9)^2)

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\frac{d}{d x} (\frac{5 x ^{4} + 3 x ^{3}}{4 x - 9})

\frac{60 x ^{4} - 156 x ^{3} - 81 x ^{2}}{( 4 x - 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5 x ^{4} + 3 x ^{3}}{4 x - 9})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 5 x ^{4} + 3 x ^{3} ) ( 4 x - 9 ) - \frac{d}{d x} ( 4 x - 9 ) ( 5 x ^{4} + 3 x ^{3} )}{( 4 x - 9 ) ^{2}}

\frac{d}{d x} (5 x^{4} + 3 x^{3}) = 20 x^{3} + 9 x^{2}

\frac{d}{d x} (4 x - 9) = 4

= \frac{( 20 x ^{3} + 9 x ^{2} ) ( 4 x - 9 ) - 4 ( 5 x ^{4} + 3 x ^{3} )}{( 4 x - 9 ) ^{2}}

Expand

(20 x^{3} + 9 x^{2}) (4 x - 9) - 4 (5 x^{4} + 3 x^{3}) : 60 x^{4} - 156 x^{3} - 81 x^{2}

= \frac{60 x ^{4} - 156 x ^{3} - 81 x ^{2}}{( 4 x - 9 ) ^{2}}

e^{x} sin (x)

\int \frac{5 - e ^{x}}{e ^{2 x}} d x

\int \frac{1}{( 64 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{220 x}{( x + 2 ) ^{2}}

\int (1 - \frac{3}{y} + x) d y

What is the derivative of (5x^4+3x^3/(4x-9)) ?
The derivative of (5x^4+3x^3/(4x-9)) is (60x^4-156x^3-81x^2)/((4x-9)^2)
What is the first derivative of (5x^4+3x^3/(4x-9)) ?
The first derivative of (5x^4+3x^3/(4x-9)) is (60x^4-156x^3-81x^2)/((4x-9)^2)