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

The derivative of f(t)=(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 f(t)=(3t-1)^4(2t+1)^{-3} ?

The first derivative of f(t)=(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

f (t) = (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}}

x \to 0 lim (\frac{tan ( k x )}{x})

\int z^{3} e^{z} d z

\frac{d}{d x} (arctan (ln (x^{2})))

\frac{d}{d x} (ln (2 x))

\int \frac{8}{x + 1} d x

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