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

The answer to derivative of f(x)=(3t-1)^4(2t+1)^{-3} is (dt)/(dx)=0

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

derivative of

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

\frac{d t}{d x} = 0

Solution steps

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

Treat

t

t (x)

Differentiate both sides:

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

Isolate

\frac{d t}{d x} : \frac{d t}{d x} = 0

\frac{d t}{d x} = 0

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

n = 1 \sum \infty \frac{5 n}{4 n + 7}

\frac{8}{( s ^{2} + 1 ) ( s ^{2} - 1 )}

\int \frac{a ^{2} - x ^{2}}{x ^{4}} d x

What is derivative of f(x)=(3t-1)^4(2t+1)^{-3} ?
The answer to derivative of f(x)=(3t-1)^4(2t+1)^{-3} is (dt)/(dx)=0