What is the derivative of f(t)=2t^3+t ?

The derivative of f(t)=2t^3+t is 6t^2+1

What is the first derivative of f(t)=2t^3+t ?

The first derivative of f(t)=2t^3+t is 6t^2+1

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derivative of

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

6 t^{2} + 1

Solution steps

\frac{d}{d t} (2 t^{3} + t)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d t} (2 t^{3}) = 6 t^{2}

\frac{d t}{d t} = 1

= 6 t^{2} + 1

\frac{\partial}{\partial x} (\frac{x y}{x + y})

f (x) = ln (6 - x^{2})

\frac{\partial}{\partial x} (x cos (y) - y sin (x))

y = \frac{2 x}{x ^{2} + 1}, (- 1, - 1)

x \to 1 lim (\frac{( 5 - 5 x )}{1 - x})

What is the derivative of f(t)=2t^3+t ?
The derivative of f(t)=2t^3+t is 6t^2+1
What is the first derivative of f(t)=2t^3+t ?
The first derivative of f(t)=2t^3+t is 6t^2+1