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

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

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

derivative of

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

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

Solution steps

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

Treat

t

t (x)

Differentiate both sides:

0 = - 3 \frac{d t}{d x} t^{4 t - 4} + 4 ln (t) \frac{d t}{d x} t^{4 t - 3} + 12 ln (t) \frac{d t}{d x} t^{4 t - 5} + 4 \frac{d t}{d x} t^{4 t - 3} + 12 \frac{d t}{d x} t^{4 t - 5} - 15 \frac{d t}{d x} t^{4 t - 6}

Isolate

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

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

\frac{\partial}{\partial x} (3 sec^{2} (2 x + 3 y))

\int \frac{e ^{x}}{4 - e ^{x}} d x

x \to 0 lim (\frac{sin ( x ) - 2 x}{x})

\frac{d}{d x} (\frac{x}{x - 1})

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

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