Cute Proofs: The Product Rule for Derivatives.

November 3, 2010

$\mbox{\tiny Two functions (solid blue and red) make the dotted purple function when multiplied!}$

There’s always a few calculus students who make the error of trying to take the derivative of $fg$ and get $f'g'$.  Of course, we know that this is not true in general, and the product rule for derivatives is as follows:

Theorem (Product Rule).  If $f, g$ are differentiable, then $(fg)' = f'g + g'f$

Given this formula, it’s a nice exercise for students to find out for which functions it is true that $(fg)' = f'g'$

Anyway, the proof of this theorem is not too difficult, even for calculus students, so I write it out for anyone interested.  This proof is really a "follow your nose" one with only one tricky part.  Let’s do it!

Proof.  We want to find $(fg)'$, so let’s set up the normal difference quotient.

$(fg)'(x) = \lim_{h\rightarrow 0}\frac{f(x + h)g(x+h) - f(x)g(x)}{h}$

Now the tricky part: we add and subtract $f(x)g(x)$ on the top.  This is really adding 0, since we’re going to add and subtract it, so this does not change our limit at all.  Thus,

$(fg)'(x) = \lim_{h\rightarrow 0}\frac{f(x + h)g(x+h) +f(x)g(x) - f(x)g(x)- f(x)g(x)}{h}$

$= \lim_{h\rightarrow 0}\frac{f(x)[g(x+h) - g(x)] + g(x+h)[f(x+h)- f(x)]}{h}$

$= \left[ \lim_{h\rightarrow 0} f(x)\right]\lim_{h\rightarrow 0}\frac{[g(x+h) - g(x)]}{h}$

$+ \left[\lim_{h\rightarrow 0}g(x+h)\right]\frac{[f(x+h)- f(x)]}{h}$

We can split up these limits because we know the limit of $f$ exists and we know the derivative of $g$ exists, and so if the limit of two functions exist and they are multiplied, the product of their limits is the limit of their products.  Anyway, this gives us

$= f(x)g'(x) + f'(x)g(x)$

which is exactly what we wanted to show.  $\Box$

This proof is pretty cute, I’m not gonna lie, and the only two tough parts were the product of two limits part and the initial adding and subtracting a common term part.

Aren’t you Bored to Death of that Proof, though?

While looking for a decent proof of this, I stumbled upon a really sweet exercise that said, "Use logarithms to prove a version of the product rule."  I’d never seen this done before, so I thought I’d share it.  In this case, we can only have that our functions are positive, which sort of restricts us a bit, but we note that since pretty much every function we care about in calculus can be written as the difference of two positive functions, and since $f,g$ have to be differentiable and therefore continuous for the product rule to work in general, this isn’t so bad of a restriction.

Proof.  Suppose that $h(x) = f(x)g(x)$.  This is equivalent to saying

$\ln(h(x)) = \ln(f(x)g(x)) = \ln(f(x)) + \ln(g(x))$

by the property of logs.  Now, take the derivative of both sides.

$\dfrac{1}{h(x)}\dfrac{dh}{dx} = \dfrac{1}{f(x)}\dfrac{df}{dx} + \dfrac{1}{g(x)}\dfrac{dg}{dx}$

Now we multiply both sides by $h = fg$, and we obtain

$\dfrac{dh}{dx} = g(x)\dfrac{df}{dx} + f(x)\dfrac{dg}{dx}$

which is exactly the product rule.  $\Box$

2 Responses to “Cute Proofs: The Product Rule for Derivatives.”

1. loiosu said

is interesting these but

2. loiosu said

here are possible the following forms Consider a function f(x) over an interval [x0, x1] as shown in Figure 3.3-1. The first degree Lagrange polynomial approximating f(x) is given by
The slope of the function at x0 is then

(x0) = (x0) + (x0) = [ f(x1)  f(x0)] + (x0)

Forward difference approximation is obtained when the slope of the interpolating polynomial estimates the derivative of the function at x0 as shown in Figure 3.3-2. In term of the forward difference operator 

(x0) = + (x0)

where f(xi) = f(xi+1)  f(xi).

Figure 3.3-2 Derivatives of the function at x0 and at x1.

The error for the derivative can be estimated by taking derivative of the error

E1(x) = ()

(x0) = [ (x  x0)(x  x1) ()

(x0) = () [(x  x0)(x  x1)

(x0) = () [(x  x1) + (x  x0) =  h () = O(h)

Backward difference approximation is obtained when the slope of the interpolating polynomial estimates the derivative of the function at x1 as shown in Figure 3.3-2.

(x1) = (x1) + (x1) = [ f(x1)  f(x0)] + (x1)

The error term has the form

(x1) = () [(x  x1) + (x  x0) = h () = O(h)

The error in the backward difference approximation, while having the same form as that in the forward difference approximation, has a different sign. In term of the backward difference operator 
Figure 3.3-3 Derivative of the function at x1 is estimated by a second-degree polynomial.

Let h = x1  x0 = x2  x1, the three points interpolating polynomial over this interval is

P2(x) = L2,0(x) f(x0) + L2,1(x)f(x1) + L2,2(x)f(x2)

P2(x) = f(x0) + f(x1) + f(x2)

P2(x) = [(x  x1)(x  x2)f(x0)  2(x  x0)(x  x2)f(x1) + (x  x0)(x  x1) f(x2)]

The function f(x) can be expressed in terms of its approximating polynomial with an error as

f(x) = P2(x) + E2(x)

The slope of the function at x1 is then

(xi) = + O(h)

where f(xi) = f(xi)  f(xi-1)

Central difference approximation is obtained when the slope of the interpolating polynomial estimates the derivative of the function at the midpoint x1 as shown in Figure 3.3-3.

P1(x) = f(x0) + f(x1)

Let h = (x1  x0), then

P1(x) = [(x1  x) f(x0) + (x  x0) f(x1)]

Figure 3.3-1 Approximating by first-degree polynomial with error E1(x).

The function f(x) can be expressed in terms of its approximating polynomial with an error as

f(x) = P1(x) + E1(x) (x1) = (x1) + (x1) = [ f(x2)  f(x0)] + (x1)

The error term in this difference approximation is

(x1) = [ (x  x0)(x  x1)(x  x2) ()
(x1) =  h2 () = O(h2)

The three points approximation is accurate to O(h2). In term of the central difference operator 

(x1) = [f(x1 + ) + f(x1  )] + O(h2)

where the central difference operator  is defined as

f(xi) = f(xi + )  (xi  )

f(x1 + ) + f(x1  ) = f(x1 + h)  f(x1) + f(x1)  f(x1  h) = f(x2)  f(x0)

Finite difference approximation of higher order derivatives can also be obtained(x) = P2(x) + E2(x)

The second derivative of the function at x1 is then

(x1) = (x1) + (x1)

After some algebra

(x1) = [f(x0)  2f(x1) + f(x2)] + (x1)

The error term can be evaluated to yield

(x1) =  h2 () = O(h2)

The central difference can also be written in terms of the central difference operator 

(x1) = + O(h2)
where

2f(x1) = [f(x1)] = [ f(x1 + )  f(x1  )] = f(x2)  f(x1)  [f(x1)  f(x0)]
So we have INTEGRATION , Prof. Orasanu

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