How can students self-check derivatives?

Question

It is a good thing for students to self-check their work.

The results of some calculations can be checked easily. For example, the solutions to an equation can be substituted back into the original to see if they really work, and integrals can be differentiated to see if it comes to the original function.

In having trouble coming up with effective ways to self-check derivatives.

I am not looking for general advice like "look for minus-sign errors", but methods that will indicate that a mistake has occurred somewhere. Note also I am not talking about differentiation by first principles using limits, but using the rules of differentiation (powers, product rule etc). I would also prefer at least some things that don't require the use of a graphics calculator/computer algebra system.

Answer 1

As a student currently taking calculus, something I do is select a couple of arbitrary points, and compare my derivative at those points with a small secant:

$f'(a) \approx \frac{f(a+h) - f(a)}{h}$

To avoid large error, I just make sure I'm not checking near any points where the derivative might be changing very quickly (eg. near 0 of $\frac{1}{x}$).

Usually, with an $h$ of 0.01, the error is extremely small, and after checking at a couple of points, I'm confident I've got it right.

What's nice is that this strategy can be extended to higher order derivatives too!

$f''(a) \approx \frac{f(a+x) - 2f(a) + f(a-x)}{x^2}$

EDIT:

For example, upon being asked to differentiate:

$f(x)=\frac{x^2 + 2}{3x - 5}$

Let's say I come up with the correct answer of:

$f'(x)=\frac{3x^2 - 10x-6}{(3x-5)^2}$

Now I want to check my answer.

First, upon analyzing $f(x)$, I see that there's a vertical asymptote at $x=5/3$. I'll want to be staying away from that region, so I decide to check the value of the derivative at $x=0$ and $x=10$.

Plugging in these values into the derivative function above I came up with:

$f'(0) = \frac{-6}{25} = -0.24$

$f'(10) = \frac{194}{625} = 0.3104$

Now I will approximate my derivative using:

$f'(x) \approx \frac{f(x+0.01) - f(x)}{0.01}$

And plugging in the values, I get:

$f'(0) \approx -0.2434607645875$

$f'(10) \approx 0.3104274870155$

Obviously, the approximations are extremely close to what my derivative function is outputting, so I'm confident I differentiated correctly.

Answer 2

One way (which I strongly encourage students to use), is to plot the original function and the derivative, and see if it makes sense in terms of increasing/decreasing behavior and concavity. This also adds a good conceptual reinforcement to what would otherwise be a mechanical problem.

A very simple example would be $f(x)=\frac{1}{3}x^3−x$. I calculate $f′(x)=x^2$. I plot $f(x)$ and $f′(x)$. Hmm. Looks like $f(x)$ is decreasing on $(−1,1)$ while $f′(x)>0$ there. Must have made a mistake. Go back and check my derivative calculation. Ah, should actually have $f′(x)=x^2−1$. Plot both. Looks good now!

Answer 3

One method, which is probably not always applicable, would be to calculate the derivative using multiple different methods. If all methods give the same answer, you've probably got it right. If not, then you've definitely got at least one of them wrong, so keep trying till all of methods give the same answer. Example:

$$\frac{d}{dx}\frac 1 {x^2}=\ ?$$

Chain rule 1:

$$= \frac{d}{dx}\frac 1 {(x^2)}=\left(\frac{d}{dx}x^2\right)\left(\frac{-1}{(x^2)^2}\right)=\frac{-2x}{x^4}=\frac{-2}{x^3}$$

Chain rule 2:

$$= \frac{d}{dx}\left(\frac 1 x\right)^2=\left(\frac{d}{dx}\frac 1 x\right)\left(2\frac 1 x\right)=\frac{-2}{x^3}$$

Quotient rule:

$$= \frac{d}{dx}\frac 1 {x^2}=\frac{0\cdot x^2-2x\cdot1}{(x^2)^2}=\frac{-2}{x^3}$$

Product rule:

$$= \frac{d}{dx} \frac 1 x \frac 1 x=\frac 1 x\frac{-1}{x^2}+\frac{-1}{x^2}\frac 1 x=\frac{-2}{x^3}$$

I still do this sometimes with complicated functions, if I get odd results and suspect I might have dropped a minus sign or something.

Answer 4

They can always check that the value of their computed derivative at a few (preferably easy) points makes sense. This is exactly how I initially attempt to convince Calc I students that $$\frac{d}{dx}e^x \neq x \, e^{x-1}.$$ Such students typically know that the graph of $y=e^x$ has positive slope at all points but the value of the purported derivative is zero at the origin.

Answer 5

I teach students to graph the numerical derivative of the original function as well as what they think the derivative is. If the student is correct, the two graphs should be identical.

We use the TI-Nspire CX, which puts each graph in a different color and makes it pretty easy to see if the graphs overlap. I also spend time on some problem functions, where the numeric derivative is not graphed as expected or where it is easily misunderstood. Students soon learn that the graphic calculator is a great aid but is not to be completely trusted.

Answer 6

Well, how about they integrate the answer to see if they get back where they started modulo a constant.

More seriously though, if the derivative is correct then we can see some things in terms of even and odd functions. The derivative of an odd function is even. The derivative of an even function is odd. Beyond that, I suppose some CAS help is the inevitable, but irrelevant, answer.

Other ideas: sometimes there is more than one way to differentiate. If they can compare product to quotient, or logarithmic to other methods that would also provide fairly disparate paths to an answer.

So, in summary, integrals are easy as differentiation to check. But, differentiation is only as safe as the student's algebra is strong. In practice, I see weakness differentiation as a common problem in the students who barely make it through the calculus sequence.