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262 Mastering Technical Mathematics<br />

DERIVATIVE OF QUOTIENT OF TWO FUNCTIONS<br />

Let f and g be two functions of a variable x, and define f/g [ f (x)]/[g (x)]. Then<br />

d( f /g)/dx [g · (df/dx) f · (dg/dx)]/g 2<br />

where g 2 g (x) · g (x), not to be confused with d 2 g/dg 2 . You might find it easier to<br />

write it like this:<br />

( f/g) (g · f f · g)/g 2<br />

RECIPROCAL DERIVATIVES<br />

Suppose that f is a function, and x and y are variables such that y f (x). The following<br />

formulas are both true as long as the derivatives exist for all values of the variables:<br />

dy/dx 1/(dx/dy)<br />

dx/dy 1/(dy/dx)<br />

DERIVATIVE OF FUNCTION RAISED TO A POWER<br />

Let f be a function of a variable x. Suppose n is a positive whole number. Then you can<br />

find the derivative of f (x) raised to the nth power by using this formula:<br />

d(f n )/dx n(f n1 ) · df/dx<br />

where f n denotes f multiplied by itself n times, not to be confused with the nth derivative.<br />

You might find it easier to write the formula like this:<br />

( f n )n( f n1 ) · f<br />

CHAIN RULE<br />

Suppose that fand g are both functions of a variable x. The derivative of the composite<br />

function (that is, f of g of x), can be found this way:<br />

[ f(g)] f(g) · g<br />

This is sometimes called by a real tongue-and-brain-twister of a name: the “formula for<br />

the derivative of a function of a function.” You must first find the derivative of “f of g<br />

of x” and then multiply the result by the derivative of “g of x.”

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