1.4 – Differentiation Using Limits of Difference Quotients  Secant line 𝑥 𝑥+ℎ ℎ 𝑥,𝑓 𝑥 𝑥,𝑓 𝑥+ℎ 𝑓 𝑥+ℎ 𝑓 𝑥 𝑦=𝑓 𝑥 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. 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑹𝒂𝒕𝒆 𝒐𝒇 𝑪𝒉𝒂𝒏𝒈𝒆= 𝒇 𝒙+𝒉 −𝒇 𝒙 𝒉 𝑺𝒍𝒐𝒑𝒆 𝒐𝒇 𝒕𝒉𝒆 𝑺𝒆𝒄𝒂𝒏𝒕 𝒍𝒊𝒏𝒆= 𝒇 𝒙+𝒉 −𝒇 𝒙 𝒉

1.4 – Differentiation Using Limits of Difference Quotients Instantaneous Rate of Change and the Slope of a Tangent Line.  Secant line 𝑥 𝑥+ℎ ℎ 𝑥,𝑓 𝑥 𝑥,𝑓 𝑥+ℎ 𝑓 𝑥+ℎ 𝑓 𝑥 𝑦=𝑓 𝑥 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. 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ

1.4 – Differentiation Using Limits of Difference Quotients Differentiation is the process used to develop the derivative. Differentiating a function will create the derivative. Slope of a Tangent Line: http://webspace.ship.edu/msrenault/GeoGebraCalculus/derivative_at_a_point.html

1.4 – Differentiation Using Limits of Difference Quotients Instantaneous Rate of Change / the Slope of a Tangent Line at a Point 𝑓 𝑥 =−2 𝑥 2 +4 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 −2 𝑥+ℎ 2 +4− −2 𝑥 2 +4 ℎ = 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 −2 𝑥 2 +2𝑥ℎ+ ℎ 2 +4− −2 1 2 +4 ℎ 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 −2 𝑥 2 −4𝑥ℎ−2 ℎ 2 +4+2 𝑥 2 −4 ℎ 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 −4𝑥ℎ−2 ℎ 2 ℎ 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 −4𝑥−2ℎ 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 ℎ −4𝑥−2ℎ ℎ 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)=−4𝑥

1.4 – Differentiation Using Limits of Difference Quotients Find (a) the slope of the tangent line at the given points and (b) the equation of the tangent line at the point. 𝑓 𝑥 =−2 𝑥 2 +4 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)=−4𝑥 1, 2 −3, −14 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)=−4𝑥 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)=−4𝑥 𝑚 𝑡𝑎𝑛 =𝑓′(1)=−4 1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ −3 =−4 −3 𝑚 𝑡𝑎𝑛 =𝑓′(1)=−4 𝑚 𝑡𝑎𝑛 = 𝑓 ′ −3 =12 𝑃𝑜𝑖𝑛𝑡−𝑆𝑙𝑜𝑝𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 𝑃𝑜𝑖𝑛𝑡−𝑆𝑙𝑜𝑝𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 𝑦−2=−4 𝑥−1 𝑦− −14 =12 𝑥− −3 𝑦−2=−4𝑥+4 𝑦+14=12𝑥+36 𝑦=−4𝑥+6 𝑦=12𝑥+22

1.4 – Differentiation Using Limits of Difference Quotients Find (a) the slope of the tangent line at the given points and (b) the equation of the tangent line at the point. 𝑓 𝑥 =−2 𝑥 2 +4 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)=−4𝑥 𝑥=2 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)=−4𝑥 𝑃𝑜𝑖𝑛𝑡−𝑆𝑙𝑜𝑝𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 𝑚 𝑡𝑎𝑛 =𝑓′(2)=−4 2 𝑦− −4 =−8 𝑥−2 𝑚 𝑡𝑎𝑛 =𝑓′(2)=−8 𝑦+4=−8𝑥+16 𝑦=−8𝑥+12 𝑓 𝑥 =−2 𝑥 2 +4 𝑓 2 =−2 2 2 +4 𝑓 2 =−4 2, −4

1.4 – Differentiation Using Limits of Difference Quotients 𝐸𝑥𝑎𝑚𝑝𝑙𝑒: 𝑓 𝑥 = 𝑥 2 +𝑥+1 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 𝑥+ℎ 2 + 𝑥+ℎ +1− 𝑥 2 +𝑥+1 ℎ = 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 𝑥 2 +2𝑥ℎ+ ℎ 2 +𝑥+ℎ+1− 𝑥 2 −𝑥−1 ℎ = 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 2𝑥ℎ+ ℎ 2 +ℎ ℎ = 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 ℎ 2𝑥+ℎ+1 ℎ = 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 = lim ℎ→0 2𝑥+ℎ+1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 =2𝑥+1

1.4 – Differentiation Using Limits of Difference Quotients Find (a) the slope of the tangent line at the given points and (b) the equation of the tangent line at the point. 𝑓 𝑥 = 𝑥 2 +𝑥+1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 =2𝑥+1 3, 13 −1, 1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 =2𝑥+1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 =2𝑥+1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 3 =2 3 +1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ −1 =2 −1 +1 𝑚 𝑡𝑎𝑛 =𝑓′(3)=7 𝑚 𝑡𝑎𝑛 = 𝑓 ′ −1 =−1 𝑃𝑜𝑖𝑛𝑡−𝑆𝑙𝑜𝑝𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 𝑃𝑜𝑖𝑛𝑡−𝑆𝑙𝑜𝑝𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 𝑦−13=7 𝑥−3 𝑦−1=−1 𝑥− −1 𝑦−13=7𝑥−21 𝑦−1=−𝑥−1 𝑦=7𝑥−8 𝑦=−𝑥

1.4 – Differentiation Using Limits of Difference Quotients Find (a) the slope of the tangent line at the given points and (b) the equation of the tangent line at the point. 𝑓 𝑥 = 𝑥 2 +𝑥+1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 =2𝑥+1 𝑥=5 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 =2𝑥+1 𝑃𝑜𝑖𝑛𝑡−𝑆𝑙𝑜𝑝𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 5 =2 5 +1 𝑦− 31 =11 𝑥−5 𝑚 𝑡𝑎𝑛 =𝑓′(5)=11 𝑦−31=11𝑥−55 𝑓 𝑥 = 𝑥 2 +𝑥+1 𝑦=11𝑥−24 𝑓 5 = 5 2 + 5 +1 𝑓 5 =31 5, 31

1.4 – Differentiation Using Limits of Difference Quotients 𝑓 𝑥 = 9 𝑥 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 −9ℎ ℎ𝑥 𝑥+ℎ 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 𝑓 𝑥+ℎ −𝑓 𝑥 ℎ = 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 9 𝑥+ℎ − 9 𝑥 ℎ 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 −9 𝑥 𝑥+ℎ 𝐿𝐶𝐷:𝑥 𝑥+ℎ 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= −9 𝑥 𝑥 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 𝑥 𝑥 ∙ 9 𝑥+ℎ − 9 𝑥 ∙ 𝑥+ℎ 𝑥+ℎ ℎ 𝑓′ 𝑥 = −9 𝑥 2 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 9𝑥 𝑥 𝑥+ℎ − 9𝑥+9ℎ 𝑥 𝑥+ℎ ℎ 𝑚 𝑡𝑎𝑛 =𝑓′(𝑥)= lim ℎ→0 9𝑥−9𝑥−9ℎ 𝑥 𝑥+ℎ ℎ

1.4 – Differentiation Using Limits of Difference Quotients Find (a) the slope of the tangent line at the given points and (b) the equation of the tangent line at the point. 𝑓 𝑥 = 9 𝑥 𝑓′ 𝑥 = −9 𝑥 2 4, 9 4 𝑃𝑜𝑖𝑛𝑡−𝑆𝑙𝑜𝑝𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 𝑥 =− 9 𝑥 2 𝑦− 9 4 =− 9 16 𝑥−4 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 4 =− 9 4 2 𝑦− 9 4 =− 9 16 𝑥+ 9 4 𝑚 𝑡𝑎𝑛 = 𝑓 ′ 4 =− 9 16 𝑦=− 9 16 𝑥+ 18 4 𝑦=− 9 16 𝑥+ 9 2

1.4 – Differentiation Using Limits of Difference Quotients Worksheet Problems Find f’(x) for each function. 1) 𝑓 𝑥 =7𝑥+32 2) 𝑓 𝑥 = 2𝑥 2 +3𝑥

1.4 – Differentiation Using Limits of Difference Quotients Worksheet Problems Find f’(x) for each function. 3) 𝑓 𝑥 = 𝑥 3 4) 𝑓 𝑥 = 𝑥−𝑥 2

1.4 – Differentiation Using Limits of Difference Quotients Worksheet Problems Find f’(x) for each function. 5) 𝑓 𝑥 = 1 𝑥 6) 𝑓 𝑥 = 1 𝑥+2

1.4 – Differentiation Using Limits of Difference Quotients