What is the derivative of (4x^4-7x/(x^3-8)) ?

The derivative of (4x^4-7x/(x^3-8)) is (4x^6-114x^3+56)/((x^3-8)^2)

What is the first derivative of (4x^4-7x/(x^3-8)) ?

The first derivative of (4x^4-7x/(x^3-8)) is (4x^6-114x^3+56)/((x^3-8)^2)

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\frac{d}{d x} (\frac{4 x ^{4} - 7 x}{x ^{3} - 8})

\frac{4 x ^{6} - 114 x ^{3} + 56}{( x ^{3} - 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x ^{4} - 7 x}{x ^{3} - 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 x ^{4} - 7 x ) ( x ^{3} - 8 ) - \frac{d}{d x} ( x ^{3} - 8 ) ( 4 x ^{4} - 7 x )}{( x ^{3} - 8 ) ^{2}}

\frac{d}{d x} (4 x^{4} - 7 x) = 16 x^{3} - 7

\frac{d}{d x} (x^{3} - 8) = 3 x^{2}

= \frac{( 16 x ^{3} - 7 ) ( x ^{3} - 8 ) - 3 x ^{2} ( 4 x ^{4} - 7 x )}{( x ^{3} - 8 ) ^{2}}

Expand

(16 x^{3} - 7) (x^{3} - 8) - 3 x^{2} (4 x^{4} - 7 x) : 4 x^{6} - 114 x^{3} + 56

= \frac{4 x ^{6} - 114 x ^{3} + 56}{( x ^{3} - 8 ) ^{2}}

f (x) = 2 x^{2} - 9 x + 10, at x = 3

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

\frac{d}{d x} (e^{(6 x - 7 x^{2})})

\int_{- 2}^{2} e^{- x^{2}} d x

y = 2 x - 2 x^{2}

What is the derivative of (4x^4-7x/(x^3-8)) ?
The derivative of (4x^4-7x/(x^3-8)) is (4x^6-114x^3+56)/((x^3-8)^2)
What is the first derivative of (4x^4-7x/(x^3-8)) ?
The first derivative of (4x^4-7x/(x^3-8)) is (4x^6-114x^3+56)/((x^3-8)^2)