3.1 –tangents and the derivative at a point the limiting value of the ratio of the change in a...
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3.1 –Tangents and the Derivative at a Point
• The limiting value of the ratio of the change in a function to the corresponding change in its independent variable.
Defn: Derivative:
• The slope of the tangent line to the graph of a function at a given point.
• The instantaneous rate of change of a function with respect to its variable.
𝑓 ′ (𝑥 )𝑟𝑒𝑎𝑑𝑠 , .𝑓 𝑝𝑟𝑖𝑚𝑒 𝑜𝑓 𝑥
3.1 –Tangents and the Derivative at a Point
3.1 –Tangents and the Derivative at a PointInstantaneous Rate of Change / the Slope of a Tangent Line at a Point
𝑓 (𝑥 )=−2 𝑥2+4 (1,2 )
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=limh→ 0
𝑓 (𝑥+h )− 𝑓 (𝑥 )(𝑥+h )− 𝑥
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=limh→ 0
−2 (1+h )2+4− (−2 (1 )2+4 )h
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (1)=−4
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=limh→ 0
−2 (1+2h+h2)+4− (−2 (1 )2+4 )h
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=limh→ 0
−2−4 h−2h2+4−2h
𝑚𝑡𝑎𝑛= 𝑓 ′ (1)=limh→ 0
𝑓 (1+h )− 𝑓 (1 )(1+h )−1
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=limh→ 0
−4h−2h2
h
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=limh→0
h (−4−2h )h
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=limh→ 0
−4−2h
3.1 –Tangents and the Derivative at a PointAlternate Definition for the Derivative at a Point
3.1 –Tangents and the Derivative at a PointInstantaneous Rate of Change / the Slope of a Tangent Line at a Point
𝑓 (𝑥 )=−2 𝑥2+4 (1,2 )
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑎)=lim𝑥→𝑎
𝑓 (𝑥 )− 𝑓 (𝑎)𝑥−𝑎
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=lim𝑥→1
−2 𝑥2+4− (−2 (1 )2+4 )𝑥−1
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (1)=−4
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=lim𝑥→1
−2 𝑥2+4−2𝑥−1
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=lim𝑥→1
−2 𝑥2+2𝑥−1
𝑚𝑡𝑎𝑛= 𝑓 ′ (1)=lim𝑥→1
𝑓 (𝑥 )− 𝑓 (1 )𝑥−1
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=lim𝑥→1
−2 (𝑥2−1 )𝑥−1
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=lim𝑥→1
−2 (𝑥+1 ) (𝑥−1 )𝑥−1
𝑚𝑡𝑎𝑛= 𝑓 ′ (1 )=lim𝑥→1
−2 (𝑥+1 )
3.2 –The Derivative of a Function
Differentiation is the process used to develop the derivative.
Differentiating a function will create the derivative.
𝐸𝑥𝑎𝑚𝑝𝑙𝑒 : 𝑓 (𝑥 )=𝑥2+𝑥+1
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=limh→ 0
𝑓 (𝑥+h )− 𝑓 (𝑥 )h
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=limh→ 0
(𝑥+h )2+ (𝑥+h )+1− (𝑥2+𝑥+1 )h
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=2 𝑥+1
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=limh→ 0
𝑥2+2 h𝑥 +h2+𝑥+h+1−𝑥2−𝑥−1h
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=limh→ 0
h (2𝑥+h+1 )h
=¿¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=limh→0
2𝑥+h+1
3.2 –The Derivative of a Function
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=limh→ 0
2 h𝑥 +h2+hh
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=lim𝑎→𝑥
𝑓 (𝑎 )− 𝑓 (𝑥 )𝑎−𝑥
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=lim𝑎→𝑥
𝑎2+𝑎+1− (𝑥2+𝑥+1 )𝑎−𝑥
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=2 𝑥+1
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=lim𝑎→𝑥
(𝑎−𝑥 ) (𝑎+𝑥+1 )𝑎−𝑥
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=lim𝑎→𝑥
(𝑎+𝑥+1 )
𝐸𝑥𝑎𝑚𝑝𝑙𝑒 : 𝑓 (𝑥 )=𝑥2+𝑥+1
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=lim𝑎→𝑥
𝑎2+𝑎+1−𝑥2−𝑥−1𝑎−𝑥
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=lim𝑎→𝑥
𝑎2+𝑎−𝑥2−𝑥𝑎− 𝑥
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=lim𝑎→𝑥
𝑎2−𝑥2+𝑎−𝑥𝑎− 𝑥
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=lim𝑎→𝑥
(𝑎+𝑥 ) (𝑎−𝑥 )+(𝑎−𝑥 )𝑎−𝑥
=¿
𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=𝑥+𝑥+1
3.2 –The Derivative of a Function