What is the derivative of (-10x+4/(3x-8)) ?

The derivative of (-10x+4/(3x-8)) is (68)/((3x-8)^2)

What is the first derivative of (-10x+4/(3x-8)) ?

The first derivative of (-10x+4/(3x-8)) is (68)/((3x-8)^2)

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\frac{d}{d x} (\frac{- 10 x + 4}{3 x - 8})

\frac{68}{( 3 x - 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{- 10 x + 4}{3 x - 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( - 10 x + 4 ) ( 3 x - 8 ) - \frac{d}{d x} ( 3 x - 8 ) ( - 10 x + 4 )}{( 3 x - 8 ) ^{2}}

\frac{d}{d x} (- 10 x + 4) = - 10

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

= \frac{( - 10 ) ( 3 x - 8 ) - 3 ( - 10 x + 4 )}{( 3 x - 8 ) ^{2}}

Simplify

\frac{( - 10 ) ( 3 x - 8 ) - 3 ( - 10 x + 4 )}{( 3 x - 8 ) ^{2}} : \frac{68}{( 3 x - 8 ) ^{2}}

= \frac{68}{( 3 x - 8 ) ^{2}}

y = x^{2} + x

\int (x + y) d x

\frac{d y}{d x}, (3 x y + 7)^{2} = 6 y

\int sec^{2} (θ) tan^{2} (θ) d θ

y^{^{''}} + 62 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the derivative of (-10x+4/(3x-8)) ?
The derivative of (-10x+4/(3x-8)) is (68)/((3x-8)^2)
What is the first derivative of (-10x+4/(3x-8)) ?
The first derivative of (-10x+4/(3x-8)) is (68)/((3x-8)^2)