1.4 – Differentiation Using Limits of Difference Quotients The Difference Quotient is used to find...

1.4 – Differentiation Using Limits of Difference Quotients

The Difference Quotient is used to find the average rate of change between two points.

The Difference Quotient also represents to slope of the secant line between two points on a curve.

Secant line

𝑥 𝑥+h

h

(𝑥 , 𝑓 (𝑥 ) )

(𝑥 , 𝑓 (𝑥+h ) )𝑓 (𝑥+h )

𝑓 (𝑥 )

𝑦= 𝑓 (𝑥 )

𝑨𝒗𝒆𝒓𝒂𝒈𝒆𝑹𝒂𝒕𝒆𝒐𝒇 𝑪𝒉𝒂𝒏𝒈𝒆=𝒇 (𝒙+𝒉 )− 𝒇 (𝒙 )

𝒉𝑺𝒍𝒐𝒑𝒆𝒐𝒇 𝒕𝒉𝒆𝑺𝒆𝒄𝒂𝒏𝒕𝒍𝒊𝒏𝒆=

𝒇 (𝒙+𝒉 )− 𝒇 (𝒙 )𝒉


The Derivative is used to find the instantaneous rate of change at any value of x.

The Derivative also represents to slope of the tangent line at any value of x.

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=limh→ 0

𝑓 (𝑥+h )− 𝑓 (𝑥 )h

Secant line

𝑥 𝑥+h

h

(𝑥 , 𝑓 (𝑥 ) )

(𝑥 , 𝑓 (𝑥+h ) )𝑓 (𝑥+h )

𝑓 (𝑥 )

𝑦= 𝑓 (𝑥 )

Instantaneous Rate of Change and the Slope of a Tangent Line.

Slope of a Tangent Line: http://webspace.ship.edu/msrenault/GeoGebraCalculus/derivative_at_a_point.html


Differentiation is the process used to develop the derivative.

Differentiating a function will create the derivative.

http://webspace.ship.edu/msrenault/GeoGebraCalculus/derivative_at_a_point.html

http://webspace.ship.edu/msrenault/GeoGebraCalculus/derivative_at_a_point.html

Instantaneous Rate of Change / the Slope of a Tangent Line at a Point

𝑓 (𝑥 )=−2 𝑥2+4


𝑓 (𝑥+h )− 𝑓 (𝑥 )h

=¿

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=limh→ 0

−2 (𝑥+h )2+4− (−2𝑥2+4 )h

=¿

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=−4 𝑥


−2 (𝑥2+2 𝑥h+h2 )+4− (−2 (1 )2+4 )h


−2𝑥2−4 𝑥 h−2h2+4+2𝑥2−4h


−4 𝑥 h−2h2

h

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=limh→0

h (−4 𝑥−2h )h


−4 𝑥−2h


𝑓 (𝑥 )=−2 𝑥2+4𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=−4 𝑥

1.4 – Differentiation Using Limits of Difference QuotientsFind (a) the slope of the tangent line at the given points and (b) the equation of the tangent line at the point.

(1 ,2 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (1)=−4 (1 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (1)=−4

𝑦−2=−4 (𝑥−1 )𝑦−2=−4 𝑥+4𝑦=−4 𝑥+6

(−3 ,−14 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (−3 )=−4 (−3 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (−3 )=12

𝑦− (−14 )=12 (𝑥− (−3 ) )𝑦+14=12𝑥+36𝑦=12 𝑥+22

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=−4 𝑥𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=−4 𝑥

𝑓 (𝑥 )=−2 𝑥2+4𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=−4 𝑥


𝑥=2

𝑚𝑡𝑎𝑛= 𝑓 ′ (2)=−4 (2 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (2)=−8𝑦− (−4 )=−8 (𝑥−2 )𝑦+4=−8 𝑥+16𝑦=−8𝑥+12𝑓 (𝑥 )=−2 𝑥2+4

𝑓 (2 )=−2 (2 )2+4𝑓 (2 )=−4

(2 ,−4 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=−4 𝑥

𝐸𝑥𝑎𝑚𝑝𝑙𝑒 : 𝑓 (𝑥 )=𝑥2+𝑥+1


𝑓 (𝑥+h )− 𝑓 (𝑥 )h

=¿


(𝑥+h )2+ (𝑥+h )+1− (𝑥2+𝑥+1 )h

=¿

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=2 𝑥+1


𝑥2+2 h𝑥 +h2+𝑥+h+1−𝑥2−𝑥−1h

=¿


h (2𝑥+h+1 )h

=¿¿

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=limh→0

2𝑥+h+1


2 h𝑥 +h2+hh

=¿



(3 ,13 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (3 )=2 (3 )+1

𝑚𝑡𝑎𝑛= 𝑓 ′ (3)=7

𝑦−13=7 (𝑥−3 )𝑦−13=7 𝑥−21𝑦=7 𝑥−8

(−1 ,1 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (−1 )=2 (−1 )+1

𝑚𝑡𝑎𝑛= 𝑓 ′ (−1 )=−1

𝑦−1=−1 (𝑥− (−1 ) )𝑦−1=−𝑥−1𝑦=−𝑥

𝑓 (𝑥 )=𝑥2+𝑥+1 𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=2 𝑥+1

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=2 𝑥+1 𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=2 𝑥+1


𝑥=5

𝑚𝑡𝑎𝑛= 𝑓 ′ (5 )=2 (5 )+1

𝑚𝑡𝑎𝑛= 𝑓 ′ (5)=11𝑦− (31 )=11 (𝑥−5 )𝑦−31=11𝑥−55𝑦=11𝑥−24

𝑓 (5 )=31(5 ,31 )

𝑓 (5 )= (5 )2+(5 )+1

𝑓 (𝑥 )=𝑥2+𝑥+1 𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=2 𝑥+1

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=2 𝑥+1

𝑓 (𝑥 )=𝑥2+𝑥+1

𝑓 (𝑥 )= 9𝑥

𝐿𝐶𝐷 :𝑥 (𝑥+h )


𝑓 (𝑥+h )− 𝑓 (𝑥 )h

=¿


9𝑥+h

−9𝑥

h

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=limh→0

𝑥𝑥∙9

𝑥+h−9𝑥∙𝑥+h𝑥+h

h𝑓 ′ (𝑥 )=−9

𝑥2


9𝑥𝑥 (𝑥+h )

−9 𝑥+9h𝑥 (𝑥+h )

h


9 𝑥−9 𝑥−9h𝑥 (𝑥+h )h


−9hh𝑥 (𝑥+h )


−9𝑥 (𝑥+h )

𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥)=−9𝑥 (𝑥 )



(4 , 94 )

𝑚𝑡𝑎𝑛= 𝑓 ′ (4 )=− 9

(4 )2

𝑦−94=−

916

(𝑥−4 )𝑚𝑡𝑎𝑛= 𝑓 ′ (𝑥 )=− 9

𝑥2

𝑓 (𝑥 )= 9𝑥

𝑓 ′ (𝑥 )=−9𝑥2

𝑚𝑡𝑎𝑛= 𝑓 ′ (4 )=− 916

𝑦−94=−

916

𝑥+94

𝑦=−916

𝑥+184

𝑦=−916

𝑥+92

1.4 – Differentiation Using Limits of Difference QuotientsWorksheet Problems

2¿ 𝑓 (𝑥 )=2𝑥2+3 𝑥1¿ 𝑓 (𝑥 )=7 𝑥+32Find f’(x) for each function.


3¿ 𝑓 (𝑥 )=𝑥3Find f’(x) for each function.

4¿ 𝑓 (𝑥 )=𝑥−𝑥2


5¿ 𝑓 (𝑥 )= 1𝑥

Find f’(x) for each function.

6¿ 𝑓 (𝑥 )= 1𝑥+2

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