Page:Calculus Made Easy.pdf/58

Then, subtracting the original $y=u\cdot v$, we have left

$dy=u\cdot dv+v\cdot du$;

and, dividing through by $dx$, we get the result:

${\frac {dy}{dx}}=u{\frac {dv}{dx}}+v{\frac {du}{dx}}$.

This shows that our instructions will be as follows: To differentiate the product of two functions, multiply each function by the differential coefficient of the other, and add together the two products so obtained.

You should note that this process amounts to the following: Treat $u$ as constant while you differentiate $v$; then treat $v$ as constant while you differentiate u; and the whole differential coefficient ${\frac {dy}{dx}}$ will be the sum of these two treatments.

Now, having found this rule, apply it to the concrete example which was considered above.

We want to differentiate the product

$(x^{2}+c)\times (ax^{4}+b)$.

Call $(x^{2}+c)=u$; and $(ax^{4}+b)=v$.

Then, by the general rule just established, we may write:

${\frac {dy}{dx}}=(x^{2}+c){\frac {d(ax^{4}+b)}{dx}}+(ax^{4}+b){\frac {d(x^{2}+c)}{dx}}$

$=(x^{2}+c)4ax^{3}+(ax^{4}+b)2x$

$=4ax^{5}+4acx^{3}+2ax^{5}+2bx$,

${\frac {dy}{dx}}=6ax^{5}+4acx^{3}+2bx$,

exactly as before.