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

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

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derivative of f(x)=(t+1)/(t^3+t-2)

derivative of

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

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

Solution steps

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

Treat

t

t (x)

Differentiate both sides:

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

Isolate

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

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

\frac{d}{d x} (7 - \frac{4}{x - 8} + \frac{2 x}{3 x + 1})

\int v d v

\frac{d}{d x} (1000 (1 8^{3} - x)^{\frac{1}{3}})

x \to 6 + lim (x^{2} - 36)

8 (e^{x} cos (x) - e^{x} sin (x))

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