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

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

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derivative of f(x)=(t^2+3)/(t^4-2t+4)

derivative of

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

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

Solution steps

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

Treat

t

t (x)

Differentiate both sides:

0 = \frac{- 2 t ^{5} \frac{d t}{d x} - 12 t ^{3} \frac{d t}{d x} - 2 t ^{2} \frac{d t}{d x} + 8 t \frac{d t}{d x} + 6 \frac{d t}{d x}}{( t ^{4} - 2 t + 4 ) ^{2}}

Isolate

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

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

x \to 0 lim (x + 2)

\int z^{31} + z^{8} d z

\int \frac{10}{x ^{0.5}} d x

y = x^{2} - 4 x, (7, 21)

4 - 3 x

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