Basic calculus question (Intuitive)

by user231094   Last Updated May 15, 2019 16:20 PM - source

So say y is a product of functions of x. For convenience, let us suppose

$y = (2x²+1) . (x+2)$

If I were to differentiate this with respect to x, I could probably use this:

$If, Y = U.V$

$→ dY/dX = U.dV + V.dU$

For U and V being functions of X

Now, doing it for the chosen example,

$y = (2x²+1) . (x+2)$

$→ dy/dx = (2x²+1)(1) + (x+2)(4x) = 6x² + 8x + 1$

And my question is, if I multiply these out first, and then take the derivative, as in:

$y = 2x³ + 4x² + x + 3$

$→ dy/dx = 6x² + 8x + 1$

Why do I get the same answer? I mean from the "basics" of derivative, we ignore really small quantities, but in these two cases, aren't the two quantities we're ignoring different for each of them?

Now I know this should make sense algebraically that I didn't change anything so I get the same answer, but why doesn't that intuit me? Is there any kind of proof or derivation (in literal sense, and not calculus derivative) that can show this out happening for a general case?

I tried to prove it and ended up deriving the formula for derivative of functions in product form. But as for the "multiply-first" kind, I could not generalize it. Any help would be thanked.



Answers 3


When taking derivative of a product we have two options.

First option is ignoring the product rule of differentiation and just multiply first and differentiate the result. This is reasonable for simple cases such as $$ y = (2x²+1) . (x+2)$$ and of course we get the correct answer $$dy/dx = 6x² + 8x + 1.$$

The second method is to apply the product rule to get derivative of the same function. We have to come up with the same answer if the product rule is to provide us with the correct result. We are not ignoring anything, we are simply using a theorem which has been proved and it saves time and effort in many cases.

You will get the same result because you are taking derivative of the same function.

You will enjoy looking at a proof of the product rule in a calculus text to really believe that it works fine.

Mohammad Riazi-Kermani
Mohammad Riazi-Kermani
May 15, 2019 15:56 PM

Intuitively, you could consider what $dy/dx$ represents geometrically.

It is the slope of the tangent line to the given curve.

The graph does not change whether you factorise the equation of the curve or write it out expanded in full.

As an aside, even if you think there is no product involved in the expanded version, there is a way of looking at the terms that show the product formula still in operation:

Take $y=2x^3$ for example, this is $y=2 \times x^3$ and you may apply the product rule to it to yield the expected result. So there is little difference between the cases considered to contain products ( $u \,.v$ ) and those considered not to contain products.

PM.
PM.
May 15, 2019 16:00 PM

Since

$$y = (2x²+1) (x+2) = 2x³ + 4x² + x + 2$$

The functions on both sides are the same not just similar, they are exactly equal. They are of the same degree and have the exact same roots. The slope of the tangent at any point is exactly the same, for any value of x, the left-hand-side will be equal to the right hand-side, as a result, the derivative of both "representations" of the function is the same.

The form $y = (2x²+1) (x+2)$ is just a factored form, not a different function. If you try to draw the curve of both the left hand side and the right hand side, you the 1 plot not 2.

$$\frac{dy}{dx}(2x²+1) (x+2)) =\frac{dy}{dx} (2x³ + 4x² + x + 2)$$

NoChance
NoChance
May 15, 2019 16:11 PM

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