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

The derivative of ((3x-4/(5x+2))^{-5}) is -(130(5x+2)^4)/((3x-4)^6)

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

The first derivative of ((3x-4/(5x+2))^{-5}) is -(130(5x+2)^4)/((3x-4)^6)

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

- \frac{130 ( 5 x + 2 ) ^{4}}{( 3 x - 4 ) ^{6}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{3 x - 4}{5 x + 2}) = \frac{26}{( 5 x + 2 ) ^{2}}

= - \frac{5}{( \frac{3 x - 4}{5 x + 2} ) ^{6}} \cdot \frac{26}{( 5 x + 2 ) ^{2}}

Simplify

- \frac{5}{( \frac{3 x - 4}{5 x + 2} ) ^{6}} \cdot \frac{26}{( 5 x + 2 ) ^{2}} : - \frac{130 ( 5 x + 2 ) ^{4}}{( 3 x - 4 ) ^{6}}

= - \frac{130 ( 5 x + 2 ) ^{4}}{( 3 x - 4 ) ^{6}}

f (x) = 5 x^{2} cos (x) + 4 x

\int_{1}^{- 1} 3 e^{x} d x

t^{2} y^{^{''}} + 2 t y^{^{'}} - 12 y = 0

f (x) = 9 x^{\frac{7}{5}} - 5 x^{2} + 1 0^{4}

3 x + sin (5 x)

What is the derivative of ((3x-4/(5x+2))^{-5}) ?
The derivative of ((3x-4/(5x+2))^{-5}) is -(130(5x+2)^4)/((3x-4)^6)
What is the first derivative of ((3x-4/(5x+2))^{-5}) ?
The first derivative of ((3x-4/(5x+2))^{-5}) is -(130(5x+2)^4)/((3x-4)^6)