Calculus: Analytic Geometry and Calculus, with Vectors

160 Functions, limits, derivatives

6 Calculate

dx(x2 + 3)(x2 - 2)

by use of the product formula. Then multiply the given factors and differentiate
the result. Make the answers agree. Hint: Look at (x2 + 3)(x2 - 2) and read
u (meaning x2 + 3) times v (meaning x2 - 2). Then apply the formula for the
derivative of uv.
7 This is another lesson on use of formulas. It is expected that persons
studying calculus are familiar with the "quadratic formula." When we want to
find the values of x for which
2x2 + 3x - 4 = 0,
we say "axe + bx + c = 0" and, without writing anything, realize that we put
a = 2 and b = 3 and c = -4 in the memorized formula

-b ± b2-4ac

2a

Then we write only

-3± 9+32

x= 4

When we use differentiation formulas, we should be equally efficient. When we

must differentiate

(1) y = (3x2 + 1)s,

we should realize that we must differentiate something of the form u" (not x's),

where u is a function of x. The formula

(2)

d

dxun=nu' 1dx

should come into our minds but should not be written. We should look at (1)

and read "y equals u to the nth power" and realize without writing anything

that u = 3x2 + 1 and n = S. We should then say "dy/dx equals n (write 5)

u (write 3x2 + 1) to the power n - I (write 4) times du/dx (write 6x)." Thus

we look at (1) and, after a little chat with ourselves, write

(3) dx = 5(3x2 + 1)4.6x

= 30x(3x2 + 1)4.

Minor modifications of this technique can be tolerated, but speed and accuracy

must be developed. Write the formula (1) and practice differentiating it as a

golfer practices putting; perfection is required.

8 Look at the calculations

y = (1 - x2), z = (1 + x2)-1

dy-

dx 1 (1- x2)-/(-2x) dxA= -(1 + x2)-2(2x)

until you see where they come from and understand them thoroughly. Nothing

is to be written.