What is the derivative of (t^2-t)(t^2+t) ?

The derivative of (t^2-t)(t^2+t) is 4t^3-2t

What is the first derivative of (t^2-t)(t^2+t) ?

The first derivative of (t^2-t)(t^2+t) is 4t^3-2t

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derivative of

(t^{2} - t) (t^{2} + t)

4 t^{3} - 2 t

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{2} - t, g = t^{2} + t

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

\frac{d}{d t} (t^{2} - t) = 2 t - 1

\frac{d}{d t} (t^{2} + t) = 2 t + 1

= (2 t - 1) (t^{2} + t) + (2 t + 1) (t^{2} - t)

Simplify

(2 t - 1) (t^{2} + t) + (2 t + 1) (t^{2} - t) : 4 t^{3} - 2 t

= 4 t^{3} - 2 t

\frac{\partial}{\partial y} (e^{3 x y})

f (x) = x - 4, (20, 40)

\frac{d y}{d x} + 5 y = 0

\frac{\partial}{\partial x} (5 + e^{- x y} cos (y))

x \to 7 lim (\frac{x ^{2} - 5 x - 14}{x - 7})

What is the derivative of (t^2-t)(t^2+t) ?
The derivative of (t^2-t)(t^2+t) is 4t^3-2t
What is the first derivative of (t^2-t)(t^2+t) ?
The first derivative of (t^2-t)(t^2+t) is 4t^3-2t