Why can't I cancel?

Question

I'm trying to differentiate the following function:

$f(x) = (5x^2-8x)e^x$

These are the steps I'm taking:

1. $[\frac {d}{dx} (5x^2-8x) \times e^x] + [\frac {d}{dx}(e^x) \times (5x^2-8x)]$

2. $[(10x-8)(e^x)] + [(e^x)(5x^2-8x)]$

3. $10x-8+5x^2-8x$

4. $5x^2+2x-8$

However, when I enter this in, it says that I'm incorrect, and that the following answer is correct:

$$e^x(5x^2+2x-8)$$

Can someone please explain to me why the $e^x$ can't be cancelled by division in step 2? I apologize if the answer is right in front of me. I know I'm just having an ID10T error.

Answer 1

The correct step from (2) to (3) should look like $$ e^x \left(10x-8\right)+e^x\left(5x^2-8x\right) \quad\Rightarrow\quad e^x \left(\left(10x-8\right)+\left(5x^2-8x\right)\right) $$ since you're pulling out an $e^x$ from both sides.

Answer 2

I always tell my students that "cancel" is not a mathematical operation.

You just saw how you can get in trouble with it above.

Consider these examples:

5/5 = 1, x/x = 1 except if x is zero x/x is undefined, (x-1)/(x-1) = 1 except when x = 1 and then (x-1)/(x-1) is undefined, So we "cancel" a number over itself and get 1 except when we don't.

5 - 5 = 0, x - x = 0, so we "cancel" a number minus itself and get zero.

$ln(e^x) = x$ except if $x \leq 0$ and then it is undefined, so we "cancel" the exponent and the logarithm and get x except when we don't.

$^\sqrt{(9 - x^2)^2} ) = 9 - x^2$ so we "cancel" the square and square root except oops!, if x is above 3 or below negative 3 whoops, the sign is wrong so wait a minute, we have to do a correction involving absolute value...

So does "cancel" mean cross out and get zero, cross out and get 1, cross out and get x, cross out and get some function inside the arentheses except maybe we need an absolute value, and does it fail for zero or one or positives or negatives or numbers above 3 and below -3, .....

Sorry, this is a shorthand that is way more trouble than it is worth.

I strongly recommend to my students to write and speak in math, and that means using mathematical operations like divide, subtract, logarithm, and when reversing mathematical operations use and speak of the concept of inverse functions. It saves a lot of time and trouble in the long run.

Answer 3

When you cancel a term like that, what you are in essence doing is dividing both sides by the term. While this is extremely useful when one side is equal to $0$ (assuming the term we are dividing by is non-zero), in this case we have $f'(x)=\text{the expression}$. Since we are trying to isolate $f'(x)$, we cannot simply cancel the term.

Hope this helps

Answer 4

In step 2 you have two terms containing $e^x$.

Take $e^x$ common we have,

$e^x [(10x-8) + (5x^2-8x)]$

Further your steps are correct. Just add $e^x$ in start of each step.

Answer 5

You wrote several expressions separately. None of them constitutes a statement by itself, and without any kind of connection, they don't combine to make any kind of statement. So basically you never said what the derivative of $f$ is, you just wrote mathematical doodles.

A good way to make a statement is to state that things are equal. Here's what your doodles look like if we put them into equations (not forgetting to say that we're looking for the derivative of $f,$ that is, $f',$ in the first place):

\begin{align} f'(x) & = \left[\frac{d}{dx}(5x^2-8x) \times e^x\right] + \left[\frac{d}{dx}(e^x) \times (5x^2-8x)\right] \\ & = [(10x-8)(e^x)] + [(e^x)(5x^2-8x)] \\ & = 10x-8+5x^2-8x \\ & = 5x^2+2x-8. \end{align}

In the middle of this, you've just claimed that $$[(10x-8)(e^x)] + [(e^x)(5x^2-8x)] = 10x-8+5x^2-8x.$$

The error should be obvious. In what sense have you ever seen this kind of "cancellation"?