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

The derivative of y=(2x+1)^3(3x+4)^{-5} is (6(2x+1)^2)/((3x+4)^5)-(15(2x+1)^3)/((3x+4)^6)

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

The first derivative of y=(2x+1)^3(3x+4)^{-5} is (6(2x+1)^2)/((3x+4)^5)-(15(2x+1)^3)/((3x+4)^6)

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derivative of

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

\frac{6 ( 2 x + 1 ) ^{2}}{( 3 x + 4 ) ^{5}} - \frac{15 ( 2 x + 1 ) ^{3}}{( 3 x + 4 ) ^{6}}

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} ((2 x + 1)^{3}) = 6 (2 x + 1)^{2}

\frac{d}{d x} ((3 x + 4)^{- 5}) = - \frac{15}{( 3 x + 4 ) ^{6}}

= 6 (2 x + 1)^{2} (3 x + 4)^{- 5} + (- \frac{15}{( 3 x + 4 ) ^{6}}) (2 x + 1)^{3}

Simplify

6 (2 x + 1)^{2} (3 x + 4)^{- 5} + (- \frac{15}{( 3 x + 4 ) ^{6}}) (2 x + 1)^{3} : \frac{6 ( 2 x + 1 ) ^{2}}{( 3 x + 4 ) ^{5}} - \frac{15 ( 2 x + 1 ) ^{3}}{( 3 x + 4 ) ^{6}}

= \frac{6 ( 2 x + 1 ) ^{2}}{( 3 x + 4 ) ^{5}} - \frac{15 ( 2 x + 1 ) ^{3}}{( 3 x + 4 ) ^{6}}

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

x^{2} d y = (x^{2} + x y + y^{2}) d x

\int 3 x^{5} d x

h \to 0 lim (- 2 x - h)

x \to 2 lim (\frac{x - 3}{x ^{2} - 4})

What is the derivative of y=(2x+1)^3(3x+4)^{-5} ?
The derivative of y=(2x+1)^3(3x+4)^{-5} is (6(2x+1)^2)/((3x+4)^5)-(15(2x+1)^3)/((3x+4)^6)
What is the first derivative of y=(2x+1)^3(3x+4)^{-5} ?
The first derivative of y=(2x+1)^3(3x+4)^{-5} is (6(2x+1)^2)/((3x+4)^5)-(15(2x+1)^3)/((3x+4)^6)