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The Product Rule

March 31st, 2009 |

Author: Merrin

Once we have learned the basic elementary functions, derived their derivatives, and also memorized their derivatives then we can learn how to differentiate more general functions constructed with these building blocks. One such construction is to take the product of two basic elementary functions. The rule for differentiating products is called the product rule.

Theorem 1: Product rule, first form. Given two functions, f(x) and g(x) which are differentiable at the point x then the derivative of their product is

$\cfrac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$

Theorem 1: Product rule, second form (less pen strokes).

$(uv)'=u'v+uv'$

Products for which we can apply the product rule are easily recognizable. Here are a few examples,

$x\sin x \qquad e^x \ln x \qquad \sqrt{x}\arcsin x \qquad \arctan x\tan x$

Here are some trickier examples

$\cfrac{\ln x}{x} \qquad 3x \qquad x^2 \qquad 0$

You may notice the first example in the list is a quotient. Well quotients are also products. Let’s rewrite it in the more suggestive form

$\left(\cfrac{1}{x}\right)(\ln x)$

The rule for the derivative of a quotient is actually just the product rule where one of the functions is considered as a reciprocal. Similarly there is no rule against one of the functions being a constant like 1 or 0. Some functions can be recast in terms of the product of two functions, for example.

$\cfrac{\sin (2x)}{2} = \sin x \cos x$

Example 1: Differentiate some examples with the product rule.

$(x \sin x )' = (x)'(\sin x) + (x)(\sin x )' = \sin x + x\cos x$

$(e^x \ln x)' = (e^x)'(\ln x ) + (e^x)(\ln x )' = e^x(\ln x + x^{-1})$

$(\sqrt{x} \arcsin x)' = (\sqrt{x})'(\arcsin x) + (\sqrt{x})(\arcsin x)'$

$(\sqrt{x} \arcsin x)' =\cfrac{\arcsin x} {2\sqrt{x}}+ \cfrac{\sqrt{x}} {\sqrt{1-x^2}}$

$(\arctan x \tan x)' = (\arctan x)'(\tan x) + (\arctan x) (\tan x)'$

$(\arctan x \tan x)'=\cfrac{\tan x}{1+x^2}+\arctan x \sec^2 x$

$(cx)' = (c)'(x)+(c)(x)'=0x+c=c$

$(0)' = (1)'(0) + (1)(0)' = (0)(0)+(1)(0) = 0$

$\left(\cfrac{ \ln x}{x} \right)' = (x^{-1})'(\ln x) + (x^{-1})(\ln x )'=-x^{-2}\ln x + x^{-1}x^{-1}$

$\left(\cfrac{ \ln x}{x} \right)' = \cfrac{1-\ln x } {x^2}$

There are no big mysteries applying the product rule. First recognize the function you want to differentiate is indeed a product of two functions which can be differentiated then apply the rule. If you can’t reach the solution is several lines then you are probably doing something wrong.

Example 2: Show the linearity property for differentiation from the product rule.

$(c_1u + c_2v)' = (c_1u)'+(c_2v)'=c_1'u+c_1u'+c_2'v+c_2v'=c_1u'+c_2v'$

$(c_1u + c_2v)' = c_1u'+c_2v'$

Example 3. Derive the product rule for a product of three functions.

$(uvw)' = (uv)'w+(uv)w'$

$(uvw)'=u'vw+uv'w+uvw'$

Each term in a multiple product gets a turn to be differentiated.

Example 4. What is the power rule for positive integers?

$(xx\cdots x)' = x'x\cdots x + xx' \cdots x + ... + xx\cdots x'= nx'x^{n-1}$

$(x^n)'=nx^{n-1}$

Now that we can apply the product rule, let’s visit a couple of proofs of the product rule.

Proof 1. We can investigate how the area of a rectangle changes which is the product of the two functions to prove the product rule another way.

$A=uv$

$A+dA=(u+du)(v+dv)=uv+udv+vdu+dudv$

The last term is negligible because it is a double differential.

$dA=udv+vdu$

$\cfrac{d}{dx} (uv)=\cfrac{du}{dx}v+ u\cfrac{dv}{dx}$

There is also a standard proof based on the definition of the limit.

Proof 2. Given y=u(x)v(x)

$\begin{array}{rcl} {\cfrac{dy}{dx} } & {=} & {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{u(x+\Delta x)v(x+\Delta x)-u(x)v(x)}{\Delta x} } \\ {} & {=} & {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{u(x+\Delta x)(v(x+\Delta x)-v(x))+v(x)(u(x+\Delta x)-u(x))}{\Delta x} } \\ {} & {=} & {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{u(x)(v(x+\Delta x)-v(x))+v(x)(u(x+\Delta x)-u(x))}{\Delta x} } \\ {} & {=} & {v(x)\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{u(x+\Delta x)-u(x)}{\Delta x} +u(x)\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{v(x+\Delta x)-v(x)}{\Delta x} } \\ {} & {=} & {u'v+uv'} \end{array}$