What is derivative of f(x)=9t^2+sqrt(t^3) ?

The answer to derivative of f(x)=9t^2+sqrt(t^3) is (dt)/(dx)=0

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derivative of f(x)=9t^2+sqrt(t^3)

derivative of

f (x) = 9 t^{2} + t^{3}

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

Solution steps

f (x) = 9 t^{2} + t^{3}

Treat

t

t (x)

Differentiate both sides:

0 = 18 t \frac{d t}{d x} + \frac{3 \frac{d t}{d x} t ^{\frac{3}{2}}}{2 t ^{2}}

Isolate

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

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

\int 4 sin^{2} (x) cos (x) d x

\frac{d}{d x} ((ln (x))^{\frac{1}{x}})

\int_{0}^{1} (x - \frac{1}{2})^{2} d x

\frac{d y}{d t} - 2 y = 3 e^{- 2 t}

\frac{d}{d x} y = (3 x + 2) (5 x^{2} + 4)

What is derivative of f(x)=9t^2+sqrt(t^3) ?
The answer to derivative of f(x)=9t^2+sqrt(t^3) is (dt)/(dx)=0