I. Computing Derivatives (slope of curve at a point) of polynomial functions.
For each of the following functions in a.-e. below perform the following three steps:
1. Compute the difference quotient f(x)=[f(x+h)-f(x)]/h
2. Simplify expression from part 1. such that h has been canceled from the denominator
3. Substitute h=0 and simplify
a. f(x)=c
b. f(x)=ax+b
c. f(x)=ax2+bx+c
d. f(x)=ax3+bx2+cx+d
e. consider f(x)=anxn+an-1xn-1+....a1x+a0, using the results from parts a. through d.,
f. find a general formula for f(x)= [f(x+h)-f(x)]/h as h→0 (steps 1 through 3 performed).
II. Show that sinθ/θ=1 as θ→0
Consider the unit circlewith θ in standard position in QI.
a. set up the inequality [(tanθ)/2]≥θ/2≥[(sinθ)/2]
b. multiply the inequality in part a. by 2/sinθ. (direction of inequalities isunchanged)
c. take the reciprocal of each term from part b. The direction of the inequality must be reversed because if a,b>0→a
d. plug in 0 for θ for cosθ only. The result should be 1≤(sinθ)/θ≤1 as θ→0
III. Show that (1-cosθ)/θ=0 as θ→0
a. multiply (1-cosθ)/θ by (1+cosθ)/(1+cosθ)
b. use trigonometric identity cos2θ+sin2θ=1 to rewrite the numerator of the expression in part a. in terms of sin2θ
c. factor the expression in part b. with one factor equal to (sinθ)/θ. (find remaining factor).
d. use the fact that (sinθ)/θ=1 as θ→0 and substitute θ=1 in the second factor (result is 0)
IV. Show that derivative of sinθ=cosθ
a. find the difference quotient f(θ)=[f(θ+h)-f(θ)]/h for f(θ)= sinθ
[use sum angle formula sin(θ+h)=sinθcosh+cosnθsinh]
b. factor sinθ out of the two terms in the numerator with sinθ in part a
c. split up the expression in part b with each term over the denominator h
d. use identities (sinh)/h=1 and (1-cosh)/h=0 as h→0 to simplify part c. to cosθ
Thus you have shown that if f(θ)=sinθ, then f(θ)=[f(θ+h)-f(θ)]/has h→0=cosθ.