AP Calculus
Mr. Manker

Difference quotient definition. Finds
derivative at a point.
𝑓 𝑥 −𝑓(𝑐)
 lim
𝑥−𝑐
𝑥→𝑐
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥

Write the equation of the tangent line to f(x)
at x = 2 if 𝑓 𝑥 = 3𝑥 2 + 7

Write the equation of the tangent line to f(x)
at x = 2 if 𝑓 𝑥 = 3𝑥 2 + 7

Use slope-intercept form: y – y1 = m(x – x1)
f(2) = 19, so (2, 19) is a point on graph
Use derivative to find slope of tan. at x = 2.

f’(x) = 6x  6(2) = 12

y – 19 = 12(x – 2)


 Is
𝑓 𝑥 = 7𝑥 − 4𝑥 + 7 increasing or
decreasing at x = 1?
2

Is 𝑓 𝑥 = 7𝑥 3 − 4𝑥 + 7 increasing or
decreasing at x = 1?
Rate of change, so find derivative at
x = 1:
2
 f‘(x) = 21𝑥 – 4
f’(1) = 17


The derivative is positive, so the
graph is increasing at x = 1.


Find f’(2) if f(x) = ln x.
We don’t know how to find the derivative of
this!!



nDeriv(ln x, x, 2) ≈ .500
To graph equation of derivative, replace x
value of 2 with variable:
nDeriv(ln x, x, x) (Good way to check your
derivatives!)

𝑓 𝑥 = 3𝑥 2 − 4𝑥 + 7
𝑓 𝑥+ℎ −𝑓(𝑥)
 lim
ℎ
ℎ→0
“forward difference quotient”
𝑓 𝑥 + ℎ − 𝑓(𝑥)
lim
ℎ→0
ℎ
3(𝑥 + ℎ)2 −4 𝑥 + ℎ + 7 − (3𝑥 2 − 4𝑥 + 7)
= lim
ℎ→0
ℎ
2
2
3𝑥 +6𝑥ℎ+3ℎ −4𝑥−4ℎ+7−3𝑥 2 +4𝑥−7
= lim
ℎ→0
ℎ
= lim 6𝑥 + 3ℎ − 4 = 6𝑥 − 4
ℎ→0





𝑑 𝑡 = 5𝑡 3 − 7𝑡 + 1
Derivative of displacement is velocity:
𝑣 𝑡 = 15𝑡 2 − 7
Derivative of velocity is acceleration:
a(t) = 30t

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