The product rule is not the first derivative rule people learn, but it might be the first non-intuitive one. \u00a0The other candidate might be the chain rule, which I’ll cover in another post.<\/p>\n
I confess that when I first saw the product rule, I wanted to doubt it. \u00a0While I agreed it must be true, it seemed to lack the elegance I expected from calculus. That was at the start of my degree, of course; a year later, I’d gotten used to it.<\/p>\n
By itself, the product rule is useful. \u00a0Moreover, several elegant techniques in later courses are based on it.<\/p>\n
The product rule is used to find the derivative of a product of functions; ie, two functions multiplied together. \u00a0One statement of it is<\/p>\n
(uv)’=uv’+u’v<\/p>\n
Let’s look at an example:<\/p>\n
Find the derivative:<\/strong> (x^2-3)sinx<\/p>\n Solution:<\/strong><\/p>\n We first need to identify the functions u and v that make the product. In this case, it seems obvious enough to decide<\/p>\n u=x^2-3<\/p>\n v=sinx<\/p>\n Then, following the product rule,<\/p>\n ((x^2-3)sinx)’=(x^2-3)(sinx)’+(x^2-3)’sinx<\/p>\n Now we take the separate derivatives as indicated:<\/p>\n ((x^2-3)sinx)’=(x^2-3)cosx + (2x)sinx<\/p>\n We expand:<\/p>\n ((x^2-3)sinx)’=x^2cosx -3cosx + 2xsinx<\/p>\n So, it turns out, by the product rule, that the derivative of (x^2-3)sinx is x^2cosx -3cosx +2xsinx.<\/p>\n Really, the product rule is kind of appealing, if you give it a chance:)<\/p>\n