Suppose u and v are functions of x that are differentiable at x=0 and that u(0)=5, u'(0)=-3, v(0)=-1, and v'(0)=2. Find the following.

Question

A) d/dx(7v-2u)

B) d/dx(u*v)

C) d/dx(v/u)

D) d/dx(u/v)

E) d/dx(3u-2v+2uv)

Answer 1

These are what I refer to as "partial problems". We don't have the exact functions for u(x) and v(x), but we do know their functions and first derivative values at one or more points along their graphs.

Let's look at part B to see how to solve these. What we'll do is use formulas and rules that you previously learned (e.g., product rule). Use the relevant rule to form an equation.

Part B:

d/dx [uv] = uv' + vu' => when x=0 we have u(0)v'(0)+v(0)u'(0) = 5*2 + (-1)(-3) = 10 + 3 = 13.

You solve the rest of them in similar fashion. Try a few to get the hang of it and check your answers below.

A. 20

B. 13

C. 7/25

D. -7

E. 15

This is a common AP Calculus exam question and a favorite type of question for many high school and college professors.

Answer 2

ASSUMING you want these derivatives evaluated at x=0, based on the given

(a) is hte linearity property of derivatives: (7v-2u)' = 7v' - 2u' = 7(2) - 2(-3) = 14 - -6 = 20

(b) is the product rule

u * v' + u' * v = 5*2 + -3 * -1 = 10 +3 = 13

(c) is the quotient rule

(u v' - u' v) / u^2 = ( 5 * 2 - -3*-3)/5^2

= (10 - 9)/25 = 1/25

(d) is quotient rule as well

( v * u' - v' * u)/ v^2

(3 - 10)/ (-1)^2 = -7

(3) 3 u ' + 2 v' + 2 ( u v' + u' v) =

3 (-3) + 2(2) + 2( 5*2 + -3 * -1) =

-9 + 4 + 2 ( 10 + 3) =

-5 + 26 =

21