# Why is $u(t)\frac{d }{dt}u(t)=\frac{1}{2}\frac{d}{dt}u(t)^2$?

If I have $$u(t)\frac{d }{dt}u(t)$$ how can I write it as $$\frac{1}{2}\frac{d}{dt}u(t)^2 \qquad \text{?}$$

Attempt: $$u(t)\frac{d }{dt}u(t)=\frac{d }{dt}\Big (u(t)u(t) \Big )=\frac{d }{dt}u(t)^2 \tag{1}$$ And just add $1/2$ to compensate for the square: $$\frac{1}{2}\frac{d}{dt}u(t)^2$$ I'm not sure of equation (1), can I really distribute the differential operator?

• Do you know the "chain rule"? – Jack Jun 29 '17 at 14:03

## 6 Answers

I think you are missing a factor of 2 in the fist equality of equation $(1)$. $$\frac{\mathrm{d}}{\mathrm{d}t} (u(t)u(t)) = \frac{\mathrm{d}u}{\mathrm{d}t} u + u \frac{\mathrm{d}u}{\mathrm{d}t} = 2\frac{\mathrm{d}u}{\mathrm{d}t} u$$ by the rule for the derivation of a product, after which your equality follows.

The chain rule states that $$[f(g(t))]' = f'(g(t))\cdot g'(t).$$

Now set $f(t) = t^2$ whose derivative is $f'(t) = 2t$. Then we have that $f(u(t)) = u(t)^2$. Now apply the chain rule:

$$[u(t)^2]' = [f(u(t))]' = f'(u(t))\cdot u'(t) = 2u(t)\cdot u'(t).$$ Hence by dividing by $2$, we get that $$\frac12 [u(t)^2]' = u(t)\cdot u'(t),$$ or written with Leibniz notation $$\frac12\frac{\mathrm d}{\mathrm d t}[u(t)^2] = u(t)\cdot \frac{\mathrm d}{\mathrm d t} u(t).$$

No, you can't just distribute across the differentiation operator. Try this: write $\frac{d}{dx}u^2(x)$ as $\frac{d}{dx}u(x)u(x)$, and then use the product rule.

By the chain rule $$\frac{d}{dt} u(t)^2 = 2 u(t) \frac{d}{dt} u(t),$$ then divide by 2.

It boils down to the chain rule: Take the right hand side $$\frac{1}{2}\frac{d}{dt}u(t)^2\stackrel{\text{chain rule}}{=}\frac{1}{2}2u(t)u'(t)=u(t)u'(t)$$

You want to use the chain rule: view $u(t)^2$ as f(u(t)) where $f(t) = t^2$. Then $$\frac{d}{dt} (u(t)^2) = \frac{d}{dt}f(u(t)) = f'(u(t)) u'(t) = 2 u(t) u'(t).$$ Divide both ends of the equation by 2 to get your answer.