Thread: Is there a way to check to see that you computed the derivative correctly?

1. Is there a way to check to see that you computed the derivative correctly?

So, I have a final exam coming up. I seem to understand and have grasped the concepts of finding derivatives quite well, but I do make mistakes here and there that cost me marks. I just wanted to know, is there any way to know that you have computed the derivative correctly that is not too complex for someone only in high school?

Like for example: The derivative of $\displaystyle x^2$ is $\displaystyle 2x$, how would I be able to know for sure that I did this correctly, even though I know this is the derivative for sure but for the more complex functions?

Thanks in advance.

2. Hey Mondreus, I wasn't talking about online derivative calculators (great tool, nonetheless), but I don't we'll be allowed to use this for the exam.

3. Yes - integrate your answer.

4. Originally Posted by TheCoffeeMachine
Yes - integrate your answer.
Hmm. While that will work, modulo an arbitrary constant, it's probably the case that integration is not yet available to the OP'er.

My recommendation: drill, drill, and drill. The differentiation rules are pretty straight-forward. It's a matter of paying attention to details.

One way to get good at math is to work in a high-energy physics lab. With 10,000 V cables running around, if you make a mistake, you're dead. Alternatively, pretend someone's holding a gun to your temple and will pull the trigger if you make a mistake.

5. If you're differentiating something like x^2 (or any function really.) Find the derivitive at a point say x=2.

To check your answer, use the equation $\displaystyle \frac{F(b)-F(a)}{b-a}$
Where $\displaystyle a<2<b$ and b,a are fairly close to 2.

If $\displaystyle \displaystyle \frac{F(b)-F(a)}{b-a} \approx \frac{d}{dx}f(x)$ at x=2
Then you have the right answer

F(b)-F(a)/b-a is the average slope on an interval, so if your interval is sufficently small then it will aprroximate your slope at that point (Which is, for all practical purposes, the definition of a derivative. Only the interval is infinitly small).

6. Originally Posted by Ackbeet
Hmm. While that will work, modulo an arbitrary constant, it's probably the case that integration is not yet available to the OP'er.

My recommendation: drill, drill, and drill. The differentiation rules are pretty straight-forward. It's a matter of paying attention to details.

One way to get good at math is to work in a high-energy physics lab. With 10,000 V cables running around, if you make a mistake, you're dead. Alternatively, pretend someone's holding a gun to your temple and will pull the trigger if you make a mistake.
Hey Ackbeet. You are correct, in that I do not have the skill to integrate functions as of yet. I do practice often, and I would love to work in a physics lab, but that option is not open to me at this time.

Do you have any other tips aside from practice? I do practice, but I still make the simple mistake here and there and feel really stupid, thereafter.

7. Originally Posted by integral
If you're differentiating something like x^2 (or any function really.) Find the derivitive at a point say x=2.

To check your answer, use the equation $\displaystyle \frac{F(b)-F(a)}{b-a}$
Where $\displaystyle a<2<b$ and b,a are fairly close to 2.

If $\displaystyle \displaystyle \frac{F(b)-F(a)}{b-a} \approx \frac{d}{dx}f(x)$ at x=2
Then you have the right answer

F(b)-F(a)/b-a is the average slope on an interval, so if your interval is sufficently small then it will aprroximate your slope at that point (Which is, for all practical purposes, the definition of a derivative. Only the interval is infinitly small).
Alright, so I'll try this:

I compute the derivative of x^2 to be 2x, and then I set x = 2. I'll set b = 2.01 and a = 1.9

f(1.9) = 3.61 and f(2.01) = 4.0401

Now to finally:
$\displaystyle \frac{4.0401-3.61}{2.01-1.9}$ =$\displaystyle \frac{0.4301}{0.11}$ = $\displaystyle 3.9$ and $\displaystyle f'(2) = 4$. So, that would be confirmation I computed the derivative correctly?

8. Originally Posted by Pupil
Alright, so I'll try this:

I compute the derivative of x^2 to be 2x, and then I set x = 2. I'll set b = 2.01 and a = 1.9

f(1.9) = 3.61 and f(2.01) = 4.0401

Now to finally:
$\displaystyle \frac{4.0401-3.61}{2.01-1.9}$ =$\displaystyle \frac{0.4301}{0.11}$ = $\displaystyle 3.9$ and $\displaystyle f'(2) = 4$. So, that would be confirmation I computed the derivative correctly?
That suggests that you did it correctly, but for erratic functions it might lead to some misleading results. For example, you would get a misleading result if you tried those same numbers with $\displaystyle f(x) = \sin(500x)$ because f is oscillating rapidly. I guess the thing to do would be, if your initial check suggests that your answer is wrong, try putting your a and b closer together. If putting a and b arbitrarily close together doesn't look like it is fixing anything then your answer is probably wrong.

9. Originally Posted by Pupil
Hey Ackbeet. You are correct, in that I do not have the skill to integrate functions as of yet. I do practice often, and I would love to work in a physics lab, but that option is not open to me at this time.

Do you have any other tips aside from practice? I do practice, but I still make the simple mistake here and there and feel really stupid, thereafter.
"Mistakes are the gems of learning." Every time you make a mistake, stop. Find the root cause of why you made the mistake. Then implement a reasonable strategy for preventing yourself from making that mistake again. That's one way to learn from your mistakes.

10. Originally Posted by Pupil
... how would I be able to know for sure that I did this correctly, even though I know this is the derivative for sure but for the more complex functions?

Thanks in advance.
Just in case a picture helps...

These are the main facts that you take as given...

... differentiating downwards. Then, what complicates matters is the chain rule...

... along with the product rule...

But you might find the patterns useful, for doing as well as checking... e.g., to differentiate $\displaystyle x^2 e^{-kx}$ you could do...

... or, zooming in further...

... and fill the blanks as here.

Also, from this kind of picture you can see The Coffee Machine's joke: integrating is harder than differentiating because fitting an expression into the bottom level of one of the shapes (chain rule or product rule), and then travelling up, is generally harder than fitting it into the top end and travelling down. Because the bottom level is more complex. So we normally recommend travelling down to check that we've travelled up correctly - not the other way round!

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

11. in a really vague, general sort of way, there are two kinds of derivatives:

1. basic building block functions: constant functions, powers of x, the exponential, logarithms, trigonometric functions (there are a few other less-frequently-used ones, but these get you started).

there is no real "quick-check method" for knowing you are differentiating these correctly, however, attempting to derive these solely from the definition of the derivative can help etch them in your memory. with some of these (logarithms, exponentials and trig functions) you need to know some special limits to get all the way through the definition derivation.

2. functions made out of the building blocks put together in various ways. this is usually where people encounter trouble.

easy stuff first: polynomial functions. the derivative is what is known as a linear operator: that is (af+bg)' = af' + bg' (where a,b are real numbers (constant multipliers), and f and g are functions of a real variable (typically x)). so polynomials are easy:

(a0 + a1x +....+ anx^n)' = a1 + 2a2x +....+ (nan)x^(n-1)

for many functions, you need to use the product rule (fg)' = fg' + f'g; or the quotient rule (f/g)' = (gf' - fg')/g^2 (many people, myself included, use some sort of doggerel as a mnemonic for this), or the chain rule. the chain rule is the most conceptually difficult to keep in mind:

(fog)' = (f'og)(g'), or as some people prefer to write it: dy/dx = (dy/du)(du/dx).

for example, when differentiating d/dx(sin(2x)) = 2cos(2x), it is quite common to forget to differentiate the "2x" inside the parentheses, and arrive at cos(2x) mistakenly.

some trig functions (notably tan(x) and sec(x)), can be derived from the basic trig functions (sin(x) and cos(x)) by using the quotient rule. although memorizing these is probably faster, it's still good practice.

one way you can use an online checker such as Wolfram|Alpha: Computational Knowledge Engine is to take a set of differentiation problems from your text-book, try to do them unaided (as if you were taking your test), and then compare your answers with the checker. the ones you get wrong, try to work through again, perhaps you can identify the TYPE of mistake you are likely to make (is it an incorrect sign? work a bit more slowly and carefully. did you miss an application of the chain rule? be on the look-out for "hidden functions" inside a parenthesis).

when you are preparing for a test, speed is not of the essence, getting it right is. there is no better way to build your confidence than working through examples, so that when your examination does occur, you can say: i know this...i've seen it before.

12. Originally Posted by Pupil
Alright, so I'll try this:

I compute the derivative of x^2 to be 2x, and then I set x = 2. I'll set b = 2.01 and a = 1.9

f(1.9) = 3.61 and f(2.01) = 4.0401

Now to finally:
$\displaystyle \frac{4.0401-3.61}{2.01-1.9}$ =$\displaystyle \frac{0.4301}{0.11}$ = $\displaystyle 3.9$ and $\displaystyle f'(2) = 4$. So, that would be confirmation I computed the derivative correctly?
Yes, this is the simplest way I know of to check to see if a derivative is right. You know it's right because 3.9 is close to 4.