Derivative Interpreted as a SLOPE
POLYNOMIAL CURVES
Equation of a Line (Recall Analytic Geometry)
Tangent and Normal to Plane Curves
“Point Slope Form”
The normal to a curve at point (x1, y1) is
defined to be the line through that point and
perpendicular to the tangent line there.
𝒚 − 𝒚𝟏 = 𝒎(𝒙 − 𝒙𝟏 )
“Slope Intercept Form”
Normal Line
𝒚 = 𝒎𝒙 + 𝒃
(x1, y1)
Tangent Line
m – Slope of the line or Gradient
b – y intercept
y – how far up
x – how far along
Slope of the Tangent Line, mt = y’ =
Slope of the Normal Line, mn =
𝑑𝑦
𝑑𝑥
−1
𝒎𝒕
Exercises:
A differentiable function of one real
variable is a function whose derivative exists at
each point in its domain. As a result, the graph of a
differentiable function must have a non-vertical
tangent line at each point in its domain, be
relatively smooth, and cannot contain any breaks,
bends, or cusps. (In other words, a continuous
function)
LIMIT DEFINITION
f’(x) =
𝑑𝑦 𝑓 (𝑥+ℎ)−𝑓(𝑥)
𝑑𝑥
=
ℎ
SLOPE
𝑡𝑎𝑛𝜃
=
𝑑𝑦
𝑑𝑥
=
𝑟𝑖𝑠𝑒
𝑟𝑢𝑛
1. Consider the curve 𝑦 = 𝑙𝑛(𝑥) + 3𝑥 2 ,
what is the slope of the curve at x = 1.5?
Answer:
𝟐𝟗
𝟑
2. Find the slope of the line at point (1, 3) that
will pass through the curve 𝑦 = 3𝑥 2 + 𝑥
Answer: 7
3. Find the slope and the equations of the
tangent and normal lines at point of the
curve indicated:
𝑦 = 3𝑥 2 − 2𝑥
at (1, 2)
Answer: mt = 4
−𝟏
mn =
𝟒
𝒚 = 𝟒𝒙 − 𝟐
𝒙 + 𝟒𝒚 = 𝟗
Equation of the Tangent Line
Equation of the Normal Line
Prepared by: ENGR. ERVIN R. ABOBO
4. Find slope and the equations of the tangent
and normal lines at the point of the curve
indicated:
𝑦 = (2𝑥 − 1)3
at x = 1
𝑦 = 𝑥 2 (𝑥 + 1)3
at (1, 8)
Answer: 28
Answer: mt = 6
−𝟏
mn =
𝟔
𝟔𝒙 − 𝒚 = 𝟓
𝒙 + 𝟔𝒚 = 𝟕
9. Find the slope of the curve
Equation of the Tangent Line
Equation of the Normal Line
10. At what point does the curve
3𝑥 2 − 7𝑥 + 𝑦 = 0 have a slope of 1?
Answer: x = 1, y = 4 or (1, 4)
5. Find slope and the equations of the tangent
and normal lines at the point of the curve
indicated:
𝑥 2 − 𝑥𝑦 + 2𝑦 − 2 = 0
at x = -2
11. At what point does the curve
𝑥 3 − 3𝑥 − 3𝑦 = 0 have a slope of 3?
Answer: x = 2, y =
𝟕
Answer: mt =
𝟖
−𝟖
mn =
𝟕
𝟕𝒙 − 𝟖𝒚 = −𝟏𝟎 Equation of the Tangent Line
𝟖𝒙 + 𝟏𝟒𝒚 = −𝟐𝟑 Equation of the Normal Line
𝟐
𝟐
or (2, )
𝟑
𝟑
12. What is the slope of the line tangent to the
function at x = 0?
𝑓𝑥 = (𝑥 3 − 5𝑥 + 3)[𝑠𝑖𝑛(3𝑥) − 2𝑥]
6. Find the slope and the equations of the
tangent and normal line at the point of the
curve indicated:
𝑥3
2
𝑦 =
at (a, a)
2𝑎−𝑥
13. What is the slope of the line tangent to the
graph of 𝑓𝑥 = 𝑠𝑖𝑛 (2𝑥) at x = 𝜋/2?
Answer: -2
Answer: mt = 2
−𝟏
mn =
𝟐
𝟐𝒙 − 𝒚 = 𝒂
𝒙 + 𝟐𝒚 = 𝟑𝒂
Answer: 3
Equation of the Tangent Line
Equation of the Normal Line
14. Find the slope of the following function at
x = 𝜋/4.
𝑦 = (𝑠𝑖𝑛𝑥)2 cos (𝑥)
7. Find the line tangent to 𝑦 = 𝑒 5𝑥 at x = 3
Answer: √𝟐/𝟒
Answer: 𝒚 = 𝟓𝒆𝟏𝟓 𝒙 − 𝟏𝟒𝒆𝟏𝟓
8. If the equation for a graph given is
3𝑥 6 + 5𝑥 2 −5𝑥+6
𝑥2
, find the slope of a
15. At what point does the curve
𝑥 3 − 9𝑥 − 𝑦 = 0 have a slope of 18?
Answer: x = 3, y = 0 or (3, 0)
line tangent to this graph at x = 5
Answer: 𝟏𝟒𝟗𝟗. 𝟗𝟒𝟒
Prepared by: ENGR. ERVIN R. ABOBO
DIFFERENTIATION OF
TRIGONOMETRIC FUNCTIONS
Trigonometric Identities
Prepared by: ENGR. ERVIN R. ABOBO
14. 𝑦 = 𝑥 2 𝑠𝑖𝑛(𝑥 ) + 2𝑥𝑐𝑜𝑠(𝑥 ) − 2𝑠𝑖𝑛(𝑥)
Exercises:
Find the first derivative of the following
function:
𝑦 ′ = 𝑥 2 𝑐𝑜𝑠𝑥
15. 𝑦 = 𝑡𝑎𝑛2 (𝑥 ) + 𝑙𝑛[𝑐𝑜𝑠 2 (𝑥 )]
𝑦 ′ = 2𝑡𝑎𝑛3 𝑥
𝑠𝑖𝑛(𝑥)
1. 𝑦 =
1+𝑐𝑜𝑠(𝑥)
16. 𝑦 = [1 + 𝑐𝑜𝑡(𝑥 )]3
1
𝑦′ =
1+𝑐𝑜𝑠(𝑥)
2. 𝑦 = 𝑠𝑖𝑛3 (𝑥 ) + 𝑠𝑖𝑛(𝑥 3 )
𝑦 ′ = 3𝑐𝑜𝑠(𝑥)𝑠𝑖𝑛2 (𝑥) + 3𝑥 2 𝑐𝑜𝑠(𝑥 3 )
3𝑥
3. 𝑦 = 𝑡𝑎𝑛 (𝑒 )
𝑦′ = 3𝑒 3𝑥 𝑠𝑒𝑐 2 (𝑒 3𝑥 )
17. 𝑦 = 𝑠𝑒𝑐 3(2𝑥 )
𝑥
18. 𝑦 = 𝑒 4𝑥 𝑐𝑜𝑠 (2)
19. (𝑥 − 𝑦)2 = 𝑠𝑖𝑛(𝑥 ) + 𝑠𝑖𝑛(𝑦)
1
20. 𝑦 = 9 𝑠𝑒𝑐(3𝑥)
21. 𝑦 = 𝑠𝑖𝑛[(√𝑥)(𝑙𝑛𝑥 )]
22. 𝑦 = 𝑐𝑠𝑐(𝑥 2 )
1+𝑐𝑜𝑠(𝑥)
4. 𝑦 =
𝑠𝑖𝑛(𝑥)
23. 𝑦 = 𝑒 𝑥𝑐𝑜𝑠𝑥
𝑦 ′ = −𝑐𝑠𝑐𝑥𝑐𝑜𝑡𝑥 − 𝑐𝑠𝑐 2 𝑥
24. 𝑦 =
5. 𝑦 = 𝑐𝑜𝑠 2(𝑠𝑖𝑛𝑥)
𝑦 ′ = −2𝑐𝑜𝑠(𝑥)𝑐𝑜𝑠(𝑠𝑖𝑛𝑥 )sin (𝑠𝑖𝑛𝑥)
6. 𝑦 = 𝑐𝑜𝑠(2𝑥) − 2𝑠𝑖𝑛(𝑥)
𝑦 ′ = −2𝑐𝑜𝑠(𝑥)(2𝑠𝑖𝑛𝑥 + 1)
𝑠𝑖𝑛(𝑥 )−𝑐𝑜𝑠(𝑥)
𝑠𝑖𝑛(𝑥)+𝑐𝑜𝑠(𝑥)
25. 𝑦 = 𝑠𝑖𝑛(𝑠𝑖𝑛𝑥)
26. 𝑦 = 𝑐𝑜𝑡 4 (2𝑥)
1
27. 𝑦 = 2 𝑠𝑒𝑐 2(𝑥 ) − 𝑙𝑛(𝑠𝑒𝑐𝑥)
1
7. 𝑦 = 𝑡𝑎𝑛𝑥 + 𝑡𝑎𝑛3 𝑥
3
𝑦 ′ = 𝑠𝑒𝑐 4 𝑥
1
28. 𝑦 = 𝑙𝑛(𝑐𝑜𝑠𝑥)
8. 𝑦 = 𝑐𝑜𝑠𝑥 − 𝑐𝑜𝑠3 𝑥
3
𝑦 ′ = −𝑠𝑖𝑛3 𝑥
9. 𝑦 = 𝑥𝑠𝑖𝑛(𝑥) + 𝑐𝑜𝑠(𝑥)
𝑦 ′ = 𝑥𝑐𝑜𝑠𝑥
2
10. 𝑦 = 𝑠𝑖𝑛 (√𝑥)
𝑦′ =
Find the second derivative of the following
function:
29. 𝑦 =
1
1+𝑐𝑜𝑠(𝑥)
30. 𝑦 = 𝑐𝑜𝑠(𝑙𝑛𝑥)
31. 𝑦 = 𝑥 2 𝑠𝑖𝑛(𝑥 ) + 2𝑥𝑐𝑜𝑠(𝑥 ) − 2𝑠𝑖𝑛(𝑥)
𝑠𝑖𝑛(√𝑥)𝑐𝑜𝑠(√𝑥)
(√𝑥)
1
11. 𝑦 = 𝑐𝑜𝑠(𝑥)
𝑦′ =
𝑠𝑖𝑛(1/𝑥)
𝑥2
12. 𝑦 = 𝑠𝑖𝑛3 (𝑥 ) + 𝑐𝑜𝑠 3 (𝑥 )
𝑦 ′ = 3𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥(𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥)
𝑥
𝑥
13. 𝑦 = 𝑡𝑎𝑛 2 − 𝑐𝑜𝑡 2
1
1
𝑦 ′ = 𝑠𝑒𝑐 2 (𝑥/2) + 𝑐𝑠𝑐 2 (𝑥/2)
2
2
Prepared by: ENGR. ERVIN R. ABOBO