# Thread: Derivative of a Vector Proof

1. ## Derivative of a Vector Proof

f(t) = real-valued function
u(t) = vector

I need help to prove the following:
d/dt [f(t)u(t)] = f'(t)u(t) + f(t)u'(t)

I know this is probably easy for you guys, but I haven't had a clue what I am doing for 4+ years now, so if this thing delves into some crazy series summation or something I am prob gonna cry.

I saw the proof for the dot product of two vectors, but I cannot do it here as I don't think I know what to do when a function is multiplied by a parametrization, I would assume you just parametrize the vector and then go from there, but I don't know what that would look like in a proof, or maybe treat the function as a scalar... sorry idk what I am talking about.

Thanks,
-Warren.

2. Originally Posted by Warrenx
f(t) = real-valued function
u(t) = vector

I need help to prove the following:
d/dt [f(t)u(t)] = f'(t)u(t) + f(t)u'(t)

I know this is probably easy for you guys, but I haven't had a clue what I am doing for 4+ years now, so if this thing delves into some crazy series summation or something I am prob gonna cry.

I saw the proof for the dot product of two vectors, but I cannot do it here as I don't think I know what to do when a function is multiplied by a parametrization, I would assume you just parametrize the vector and then go from there, but I don't know what that would look like in a proof, or maybe treat the function as a scalar... sorry idk what I am talking about.

Thanks,
-Warren.

Just think of it as the product rule you learned in calc 1.

3. Originally Posted by Warrenx
f(t) = real-valued function
u(t) = vector

I need help to prove the following:
d/dt [f(t)u(t)] = f'(t)u(t) + f(t)u'(t)

I know this is probably easy for you guys, but I haven't had a clue what I am doing for 4+ years now, so if this thing delves into some crazy series summation or something I am prob gonna cry.

I saw the proof for the dot product of two vectors, but I cannot do it here as I don't think I know what to do when a function is multiplied by a parametrization, I would assume you just parametrize the vector and then go from there, but I don't know what that would look like in a proof, or maybe treat the function as a scalar... sorry idk what I am talking about.

Thanks,
-Warren.

What is $\displaystyle \bold{u}$? I assume it's a vector valued function $\displaystyle \bold{u}:\mathbb{R}\to\mathbb{R}^n:t\mapsto(u_1(t) ,\cdots,u_n(t))$. If so maybe it's easier to think about it by $\displaystyle \bold{u}(t)=\sum_{j=1}^{n}u_j(t)e_j$ (where $\displaystyle e_j=(0,\cdots,\underbrace{1}_{j^{\text{th}}\text{ }place},\cdots,0)$) and then it should seem obvious that $\displaystyle f(t)\bold{u}(t)=\sum_{j=1}^{n}f(t)u_j(t)e_j$. But then how do we define the differentiation of this function? Isn't it coordinate-wise? so that $\displaystyle \frac{d}{dt}\left(f(t)\bold{u}(t)\right)=\sum_{j=1 }^{n}\frac{d}{dt}(f(t)u_j(t))e_j=\cdots$

Can you finish?