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

The derivative of (4x+3)^4(x+1)^{-3} is (16(4x+3)^3)/((x+1)^3)-(3(4x+3)^4)/((x+1)^4)

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

The first derivative of (4x+3)^4(x+1)^{-3} is (16(4x+3)^3)/((x+1)^3)-(3(4x+3)^4)/((x+1)^4)

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derivative of

(4 x + 3)^{4} (x + 1)^{- 3}

\frac{16 ( 4 x + 3 ) ^{3}}{( x + 1 ) ^{3}} - \frac{3 ( 4 x + 3 ) ^{4}}{( x + 1 ) ^{4}}

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

= 16 (4 x + 3)^{3} (x + 1)^{- 3} + (- \frac{3}{( x + 1 ) ^{4}}) (4 x + 3)^{4}

Simplify

16 (4 x + 3)^{3} (x + 1)^{- 3} + (- \frac{3}{( x + 1 ) ^{4}}) (4 x + 3)^{4} : \frac{16 ( 4 x + 3 ) ^{3}}{( x + 1 ) ^{3}} - \frac{3 ( 4 x + 3 ) ^{4}}{( x + 1 ) ^{4}}

= \frac{16 ( 4 x + 3 ) ^{3}}{( x + 1 ) ^{3}} - \frac{3 ( 4 x + 3 ) ^{4}}{( x + 1 ) ^{4}}

\frac{\partial}{\partial v} (\frac{v}{u - v})

\frac{d}{d x} (\frac{x ^{3}}{5})

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

(1, 2), (- 2, - 1)

\int \frac{3 x}{1 - x ^{4}} d x

What is the derivative of (4x+3)^4(x+1)^{-3} ?
The derivative of (4x+3)^4(x+1)^{-3} is (16(4x+3)^3)/((x+1)^3)-(3(4x+3)^4)/((x+1)^4)
What is the first derivative of (4x+3)^4(x+1)^{-3} ?
The first derivative of (4x+3)^4(x+1)^{-3} is (16(4x+3)^3)/((x+1)^3)-(3(4x+3)^4)/((x+1)^4)