Can you cancel out the derivative function - very new to calculus

Question

I'm incredibly new to calculus (I'm a high school freshman and I'm doing a summer camp that involves calculus when I've only just completed geometry). While I do enjoy the rigor of the course, it is daunting to have to self study when you barely know where to look and can barely understand math tutorials because the information and problems presented to me is so far from my normal - So I understand if my question might seem stupid or silly but I truly am so lost.

My question is - can you cancel out the derivative function - for example if you have have been given that the position of a particle is $x = 20t - 5t^3$ where $x$ is in meters and $t$ is in seconds and you are asked to find where the velocity is equal to zero, you first have to find what $\frac{\mathrm{d}x}{\mathrm{d}t}$ is (I assume). So then you get $\frac{\mathrm{d}(20t - 5t^3)}{\mathrm{d}t}$. I looked at the answer for guidance and it says the answer for that equation is $20t - 15t$ which I don't understand unless you can cancel out the differential functions and then the $t$'s. I am finding a lot of flaws with my logic but if someone could please give me a step by step answer to this without skipping my steps (because I probably could not figure out how to go from one to the other), that would be much appreciated.

Answer 1

If $ x(t) = 20 t - 5t^3$, then $x'(t) = 20 - 15t^2$.

The derivative rule for polynomials is $$ \dfrac{d}{dt} at^b = (a \times b) t^{b-1}. $$

Answer 2

In Leibniz derivative notation, which can be written as: $$\frac{d}{dt}\left(20t - 5t^3\right),$$the $d$ and $dt$ are not variables; they simply show what order derivative is with respect to what variable.

$$\frac{d}{dt}\left(20t - 5t^3\right)\ne 20t-15t=5t.$$

To find the derivative, you need to apply the power rule to the second term:

$$\frac{d}{dt}\left(20t - 5t^3\right)=\frac{d}{dt}\left(20t\right) - \frac{d}{dt}\left(5t^3\right)=20-15t^2$$.

Answer 3

Using only the definition of the derivative...

\begin{align*} \frac{\mathrm{d}f(t)}{\mathrm{d}t} &= \lim_{\Delta t \rightarrow 0} \frac{f(t+\Delta t) - f(t)}{(t+\Delta t) - t} \end{align*} (This should very much look like a "slope of a line" computation. We have the change in the output of the function divided by the change in the input to the function, $\frac{\Delta f}{\Delta t}$. We are taking the limit as the "nudge" in the input shrinks to zero, since we just want to know the rate of change of the output to change of the input, which is indicated in the derivative notation by the change from "$\Delta$"s to "$\mathrm{d}$"s.) \begin{align*} &= \lim_{\Delta t \rightarrow 0} \frac{(20(t+\Delta t) - 5(t+\Delta t)^3) - (20t - 5t^3)}{\Delta t} \\ &= \lim_{\Delta t \rightarrow 0} \frac{20t +20\Delta t - 5t^3 - 15 t^2 \Delta t - 15 t (\Delta t)^2 - 5(\Delta t)^3 - 20t + 5t^3}{\Delta t} \\ &= \lim_{\Delta t \rightarrow 0} \frac{20\Delta t - 15 t^2 \Delta t - 15 t (\Delta t)^2 - 5(\Delta t)^3 }{\Delta t} \\ &= \lim_{\Delta t \rightarrow 0} \left( 20 - 15 t^2 - 15 t \Delta t - 5(\Delta t)^2 \right) \\ &= 20 - 15 t^2 - 15 t (0)- 5(0)^2 \\ &= 20 - 15 t^2 \text{.} \end{align*} (Here we discharge the limit using the continuity of polynomials. Here, we have a polynomial in $\Delta t$. Recall that a function is continuous if (and only if) for all inputs, its value agrees with its limits approaching those inputs. So we evaluate the limit of the continuous function by evaluating the function at the limit.)