What is the derivative of (3t-1)^4(2t+1)^{-3} ?

The derivative of (3t-1)^4(2t+1)^{-3} is (12(3t-1)^3)/((2t+1)^3)-(6(3t-1)^4)/((2t+1)^4)

What is the first derivative of (3t-1)^4(2t+1)^{-3} ?

The first derivative of (3t-1)^4(2t+1)^{-3} is (12(3t-1)^3)/((2t+1)^3)-(6(3t-1)^4)/((2t+1)^4)

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derivative of

(3 t - 1)^{4} (2 t + 1)^{- 3}

\frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d t} ((3 t - 1)^{4}) = 12 (3 t - 1)^{3}

\frac{d}{d t} ((2 t + 1)^{- 3}) = - \frac{6}{( 2 t + 1 ) ^{4}}

= 12 (3 t - 1)^{3} (2 t + 1)^{- 3} + (- \frac{6}{( 2 t + 1 ) ^{4}}) (3 t - 1)^{4}

Simplify

12 (3 t - 1)^{3} (2 t + 1)^{- 3} + (- \frac{6}{( 2 t + 1 ) ^{4}}) (3 t - 1)^{4} : \frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}

= \frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}

\int (\frac{x}{1 + x ^{2}}) d x

x \to 4 + lim (\frac{1}{x - 4})

\frac{d y}{d x} = 2 x - y

x \to 2 lim (9 (ln (5 x) + 1))

\int 3 sin (t) - 2 cos (t) d t

What is the derivative of (3t-1)^4(2t+1)^{-3} ?
The derivative of (3t-1)^4(2t+1)^{-3} is (12(3t-1)^3)/((2t+1)^3)-(6(3t-1)^4)/((2t+1)^4)
What is the first derivative of (3t-1)^4(2t+1)^{-3} ?
The first derivative of (3t-1)^4(2t+1)^{-3} is (12(3t-1)^3)/((2t+1)^3)-(6(3t-1)^4)/((2t+1)^4)