What is the derivative of y=(5x+1)^4(4x+3)^{-2} ?

The derivative of y=(5x+1)^4(4x+3)^{-2} is (20(5x+1)^3)/((4x+3)^2)-(8(5x+1)^4)/((4x+3)^3)

What is the first derivative of y=(5x+1)^4(4x+3)^{-2} ?

The first derivative of y=(5x+1)^4(4x+3)^{-2} is (20(5x+1)^3)/((4x+3)^2)-(8(5x+1)^4)/((4x+3)^3)

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derivative of

y = (5 x + 1)^{4} (4 x + 3)^{- 2}

\frac{20 ( 5 x + 1 ) ^{3}}{( 4 x + 3 ) ^{2}} - \frac{8 ( 5 x + 1 ) ^{4}}{( 4 x + 3 ) ^{3}}

Solution steps

\frac{d}{d x} ((5 x + 1)^{4} (4 x + 3)^{- 2})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (5 x + 1)^{4}, g = (4 x + 3)^{- 2}

= \frac{d}{d x} ((5 x + 1)^{4}) (4 x + 3)^{- 2} + \frac{d}{d x} ((4 x + 3)^{- 2}) (5 x + 1)^{4}

\frac{d}{d x} ((5 x + 1)^{4}) = 20 (5 x + 1)^{3}

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

= 20 (5 x + 1)^{3} (4 x + 3)^{- 2} + (- \frac{8}{( 4 x + 3 ) ^{3}}) (5 x + 1)^{4}

Simplify

20 (5 x + 1)^{3} (4 x + 3)^{- 2} + (- \frac{8}{( 4 x + 3 ) ^{3}}) (5 x + 1)^{4} : \frac{20 ( 5 x + 1 ) ^{3}}{( 4 x + 3 ) ^{2}} - \frac{8 ( 5 x + 1 ) ^{4}}{( 4 x + 3 ) ^{3}}

= \frac{20 ( 5 x + 1 ) ^{3}}{( 4 x + 3 ) ^{2}} - \frac{8 ( 5 x + 1 ) ^{4}}{( 4 x + 3 ) ^{3}}

x \to 4 lim (x \cdot 5)

x \to \infty lim (x + \frac{1}{x})

\int_{2}^{10} (- 4) d x

\frac{d}{d x} (x sin (ln (x)))

n = 0 \sum \infty \frac{1}{n + 19}

What is the derivative of y=(5x+1)^4(4x+3)^{-2} ?
The derivative of y=(5x+1)^4(4x+3)^{-2} is (20(5x+1)^3)/((4x+3)^2)-(8(5x+1)^4)/((4x+3)^3)
What is the first derivative of y=(5x+1)^4(4x+3)^{-2} ?
The first derivative of y=(5x+1)^4(4x+3)^{-2} is (20(5x+1)^3)/((4x+3)^2)-(8(5x+1)^4)/((4x+3)^3)