# Thread: Chain rule, I think

1. ## Chain rule, I think

A particle moves along the x-axis with the velocity dx/dt= f(x). Show that the particle's acceleration is f(x)f '(x).

The notation looks funny. It usually is dy/dx. This time it's the derivative of x with respect to t.

Tried chain rule (where f is outside and x is inside), and the product rule.

2. Originally Posted by Truthbetold
A particle moves along the x-axis with the velocity dx/dt= f(x). Show that the particle's acceleration is f(x)f '(x).

The notation looks funny. It usually is dy/dx. This time it's the derivative of x with respect to t.

Tried chain rule (where f is outside and x is inside), and the product rule.
Hello,

you have: $\frac{dx}{dt}=f(x)$.

As you correctly pointed out the acceleration is the first derivative wrt t of the velocity:

$\frac{d^2x}{dt^2}=f'(x) \cdot \underbrace{\frac{dx}{dt}}_{\text{equals f(x)}}=f(x) \cdot f'(x)$.

3. Just curious here, but if you're differentiating x with respect to t, shouldn't it be $\frac{\,dt}{\,dx}$

As it is in your first post, $\frac{\,dx}{\,dt}=f(x)$ makes little sense.

4. Originally Posted by DivideBy0
Just curious here, but if you're differentiating x with respect to t, shouldn't it be $\frac{\,dt}{\,dx}$

As it is in your first post, $\frac{\,dx}{\,dt}=f(x)$ makes little sense.
Nope, the book says dx/dt= f(x).

5. I find that confusing

6. Originally Posted by earboth
Hello,

you have: $\frac{dx}{dt}=f(x)$.

As you correctly pointed out the acceleration is the first derivative wrt t of the velocity:

$\frac{d^2x}{dt^2}=f'(x) \cdot \underbrace{\frac{dx}{dt}}_{\text{equals f(x)}}=f(x) \cdot f'(x)$.
I don't follow.
WOuldn't the derivative of f(x) just be f'(x) and not f(x)f'(x)?
I know the notation has to do with something as you showed the difference of $\frac{d^2x}{dt^2}=f'(x)$ and $\frac{dx}{dt}$.

7. its the chain rule.

8. Originally Posted by Truthbetold
I don't follow.
WOuldn't the derivative of f(x) just be f'(x) and not f(x)f'(x)?
I know the notation has to do with something as you showed the difference of $\frac{d^2x}{dt^2}=f'(x)$ and $\frac{dx}{dt}$.
Hello,

"WOuldn't the derivative of f(x) just be f'(x)" This statement is only true if you derivate wrt x. Now the problem says that you have to derivate wrt t

As shilz222 pointed out you have to use the chain rule because x is a function of t, that means x isn't a single variable here but a function.