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πππ πππ= 1 π πππ πππ πππ= 1 π πππ πππ πππ= 1 π πππ
πππ πππ= 1 π πππ πππ πππ= 1 π πππ πππ πππ= 1 π πππ Derivative sin x, cos x 6A
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Trig Differentiation Starter: Sketch the graphs of sinx, cos x tan x
KUS objectives BAT differentiate trig functions using what we have learned so far BAT differentiate mixed trig, exponential, logarithm and other functions Starter: Sketch the graphs of sinx, cos x tan x Where are the stationary points on sinx, cos x tan x ? Sketch graphs of arcsin x, arccos x, arctan x
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The gradient function of:
Gradients I: Trig Graphs The gradient function of: y = sinx y = cosx dy/dx= cosx dy/dx= -sinx
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WB 1 Differentiate each of : a) π¦= sin 3π₯ b) π¦= sin 2 3 π₯
c) π¦= π ππ 2 π₯ d) π¦= πππ 2 π₯ e) y=cos 3π₯ 2 β8 π) π¦=π ππ3π₯ d) π¦=( cos π₯ ) 2 ππ¦ ππ₯ =3 cos 3π₯ ππ¦ ππ₯ =2 ( cos π₯ ) 1 β sin π₯ =β2 sin π₯ cos π₯ =β sin 2π₯ b) π¦=π ππ 2 3 π₯ e) π¦=cos 3π₯ 2 β8 ππ¦ ππ₯ = 2 3 πππ 2 3 π₯ ππ¦ ππ₯ =βsin 3π₯ 2 β8 Γ(6π₯) ππ¦ ππ₯ =β6π₯ sin 3π₯ 2 β8 π) π¦=(π πππ₯ ) 2 ππ¦ ππ₯ =2(π πππ₯ ) 1 (πππ π₯) ππ¦ ππ₯ =2π πππ₯πππ π₯=sin 2x
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WB 2 Differentiate each of : a) π¦= πππ (4π₯β3) b) π¦= πππ 3 π₯
c) π¦= 1 2 sin 2π₯ 2 +2π₯ d) π¦= π ππ 2 π₯ e) y= 1 2 πππ 2π₯+1 2 π) π¦= sin π₯ π) π¦=πππ β‘(4π₯β3) ππ¦ ππ₯ =β4π ππβ‘(4π₯β3) ππ¦ ππ₯ =2 sin π₯ cos π₯ 2 ππ¦ ππ₯ = sin π₯ 2 cos π₯ 2 = sin π₯ 2 π) π¦=(πππ π₯ ) 3 ππ¦ ππ₯ =3(πππ π₯ ) 2 (β sin π₯ ) e) π¦= 1 2 cos 2π₯+1 2 ππ¦ ππ₯ =β3ππ π 2 π₯ π πππ₯ ππ¦ ππ₯ =β 1 2 sin 2π₯+1 2 Γ2(2π₯+1) c) π¦= 1 2 sin 2π₯ 2 +2π₯ ππ¦ ππ₯ = 2x+1 sin 2π₯+1 2 ππ¦ ππ₯ = 1 2 cos 2π₯ 2 +2π₯ Γ(4π₯+2) ππ¦ ππ₯ = 2x+1 sin 2π₯ 2 +2π₯
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Product and quotient rules
Derivative tan x Product and quotient rules If: π¦=π‘πππ₯ ππ¦ ππ₯ =π π π 2 π₯ Then: If: π¦=π‘ππβ‘π π₯ ππ¦ ππ₯ = π β² π₯ π π π 2 π(π₯) Then:
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WB 3ab Differentiate each of : a) π¦= π₯ 2 cos π₯ b) π¦=4π₯ sin 2π₯
c) π¦= 1 2 sin π₯ cos 2π₯ d) π¦= sin 3π₯ cos π₯ a) π¦= π₯ 2 cos π₯ Use the product rule WORKING OUT SPACE ππ¦ ππ₯ = cos π₯ Γ 2π₯ + π₯ 2 Γβ sin π₯ π’= π₯ 2 π£= cos π₯ ππ¦ ππ₯ =2π₯ cos π₯ β π₯ 2 sin π₯ ππ’ ππ₯ =2π₯ ππ£ ππ₯ =β sin π₯ b) π¦= 4π₯ sin 2π₯ Use the product rule WORKING OUT SPACE ππ¦ ππ₯ = sin 2π₯ Γ π₯ Γ2 cos 2π₯ π’=4π₯ π£= sin 2π₯ ππ¦ ππ₯ =4 sin 2π₯ +8π₯ cos 2π₯ ππ’ ππ₯ =4 ππ£ ππ₯ =2 cos 2π₯
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WB 3cd Differentiate each of : a) π¦= π₯ 2 cos π₯ b) π¦=4π₯ sin 2π₯
c) π¦= 1 2 sin π₯ cos 2π₯ d) π¦= sin 3π₯ cos π₯ c) π¦= sin π₯ cos 2π₯ Use the product rule WORKING OUT SPACE π’= sin π₯ π£= 1 2 cos 2π₯ ππ¦ ππ₯ = 1 2 cos 2π₯ Γ cos π₯ + sin π₯ Γ β sin 2π₯ ππ¦ ππ₯ = 1 2 cos x cos 2π₯ β sin π₯ sin 2π₯ ππ’ ππ₯ = cos π₯ ππ£ ππ₯ =β sin 2π₯ c) π¦= sin 3π₯ cos π₯ Use the product rule WORKING OUT SPACE ππ¦ ππ₯ = cos π₯ Γ 3 cos 3π₯ + sin 3π₯ Γ β sin π₯ π’= sin 3π₯ π£= cos π₯ ππ¦ ππ₯ =3cos x cos 2π₯ β sin π₯ sin 3π₯ ππ’ ππ₯ =3 cos 3π₯ ππ£ ππ₯ =β sin π₯
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WB 4ab Differentiate each of : a) π¦= sin 10π₯ 5π₯ b) π¦= π₯ 2 cos π₯
WORKING OUT SPACE Use the quotient rule π’= sin 10π₯ π£=5π₯ ππ¦ ππ₯ = 5π₯Γ 10 cos 10π₯ β sin 10π₯ Γ5 5π₯ 2 ππ’ ππ₯ =10 cos 10π₯ ππ£ ππ₯ =5 ππ¦ ππ₯ = 10π₯ cos 10π₯ β sin 10π₯ 5 π₯ 2 b) π¦= π₯ 2 cos π₯ WORKING OUT SPACE Use the quotient rule π’= π₯ 2 π£= cos π₯ ππ¦ ππ₯ = cos π₯ Γ 2π₯ β π₯ 2 Γ β sin π₯ cos π₯ 2 ππ’ ππ₯ =2π₯ ππ£ ππ₯ =β sin π₯ ππ¦ ππ₯ = 2π₯ cos π₯ + π₯ 2 sin π₯ πππ 2 π₯
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Quotients Rule! ππ¦ ππ₯ = cos π₯ Γ cos π₯ β sin π₯ Γ β sin π₯ cos π₯ 2
WB 5 Find the derivative of tan x π¦= sin π₯ cos π₯ ππ¦ ππ₯ = π£ ππ’ ππ₯ βπ’ ππ£ ππ₯ π£ 2 Use the quotient rule WORKING OUT SPACE π’= sin π₯ π£= cos π₯ ππ¦ ππ₯ = cos π₯ Γ cos π₯ β sin π₯ Γ β sin π₯ cos π₯ 2 ππ’ ππ₯ = cos π₯ ππ£ ππ₯ =β sin π₯ ππ¦ ππ₯ = π ππ 2 π₯ + β πππ 2 π₯ πππ 2 π₯ ππ¦ ππ₯ = 1 πππ 2 π₯ = π ππ 2 π₯ π ππ₯ tan π₯ = π ππ 2 π₯
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WB 6 Differentiate each of : a) π¦=π‘π π 4 π₯ b) π¦=π₯ tan 2π₯
ππ¦ ππ₯ =4(π‘πππ₯ ) 3 (π π π 2 π₯) ππ¦ ππ₯ =4π‘π π 3 π₯π π π 2 π₯ b) π¦=π₯ π‘ππ2π₯ WORKING OUT SPACE Use the product rule ππ¦ ππ₯ =π₯Γ(2π π π 2 2π₯) +(π‘ππ2π₯)Γ1 π’=π₯ π£=π‘ππ2π₯ ππ’ ππ₯ =1 ππ£ ππ₯ =2π π π 2 2π₯ ππ¦ ππ₯ =2π₯π π π 2 2π₯+ π‘ππ2π₯
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πππ πππ= 1 π πππ πππ πππ= 1 π πππ πππ πππ= 1 π πππ
πππ πππ= 1 π πππ πππ πππ= 1 π πππ πππ πππ= 1 π πππ Derivative Mixed functions
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ππ¦ ππ₯ = π πππ₯Γ 1 π₯ β ln π₯ Γ cos π₯ sin π₯ 2
WB 7ab Differentiate each of : a) π¦= ln π₯ sin π₯ b) π¦= π π₯ sin π₯ a) π¦= ln π₯ sin π₯ ππ¦ ππ₯ = π£ ππ’ ππ₯ βπ’ ππ£ ππ₯ π£ 2 Use the quotient rule WORKING OUT SPACE ππ¦ ππ₯ = π πππ₯Γ 1 π₯ β ln π₯ Γ cos π₯ sin π₯ 2 π’=πππ₯ π£=π πππ₯ ππ’ ππ₯ = 1 π₯ ππ£ ππ₯ =πππ π₯ ππ¦ ππ₯ = π πππ₯ β π₯ ln π₯ cos π₯ π₯ π ππ 2 π₯ b) π¦= π π₯ sin π₯ Use the product rule WORKING OUT SPACE ππ¦ ππ₯ =π πππ₯Γ π π₯ + π π₯ Γ cos π₯ π’= π π₯ π£=π πππ₯ ππ’ ππ₯ = π π₯ ππ£ ππ₯ =πππ π₯ ππ¦ ππ₯ = π π₯ sin π₯ + cos π₯
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ππ¦ ππ₯ = ππΓ π ππ 2 π₯ β tan π₯ Γ 1 π₯ tan π₯ 2
WB 7cd Differentiate each of : c) π¦= π π₯ 2 sin π₯ 2 d) π¦= tan π₯ ln π₯ c) π¦= π π₯ sin π₯ Use the product rule WORKING OUT SPACE ππ¦ ππ₯ = sin π₯ 2 Γ 2π₯ π π₯ π π₯ 2 Γ cos π₯ 2 π’= π π₯ 2 π£= sin π₯ 2 ππ’ ππ₯ =2π₯ π π₯ 2 ππ£ ππ₯ = 1 2 cos π₯ 2 ππ¦ ππ₯ = π π₯ 2 2π₯ sin π₯ cos π₯ d) π¦= tan π₯ ln π₯ Use the quotient rule WORKING OUT SPACE ππ¦ ππ₯ = ππΓ π ππ 2 π₯ β tan π₯ Γ 1 π₯ tan π₯ 2 π’= tan π₯ π£= ln π₯ ππ’ ππ₯ = π ππ 2 π₯ ππ£ ππ₯ = 1 π₯ ππ¦ ππ₯ = π₯ ln π₯ π ππ 2 π₯β tan π₯ π₯ π₯ π‘ππ 2 π₯
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self-assess using: R / A / G βI am now able to ____ .
KUS objectives BAT differentiate trig functions using what we have learned so far BAT differentiate mixed trig, exponential, logarithm and other functions self-assess using: R / A / G βI am now able to ____ . To improve I need to be able to ____β
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