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MCV4U Unit # 4 – Extensions Assessment of Learning 2021/2022
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MCV4U Unit # 4 – Extensions Assessment of Learning Assessment of Learning Unit # 4 – Extensions 1 a. Find lim
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Unit # 4 – Extensions MCV4U Unit # 4 – Extensions Assessment of Learning Assessment of Learning Unit # 4 – Extensions 1 a. Find lim𝑥→0−2 sin 3𝑥 . Defend your answer. (2 marks) Consider this graph. From the graph we see the y value approaches 0, the x value approaches 0 from both left and right side. Therefore we know lim
𝑥→0−2sin3𝑥 = (−2)(0)= 0 1.b Find lim
𝑥→0−𝑐𝑜𝑠5𝑥 . Defend your answer. (2 marks) Consider this graph. From the graph we see the Y value approach 1, the x value approaches o from both the right and left side. Therefore, lim
𝑥→0−𝑐𝑜𝑠5𝑥 = (−1)(1)= 1 . Unit # 4 – Extensions 2. Find the derivative
a.𝑦 = 𝑠𝑖𝑛𝑥5 (1 mark)
𝑦′=(𝑠𝑖𝑛𝑥5)′ 𝑦′= cos 𝑥5(𝑥5)′ 𝑦′= cos 𝑥5(5𝑥5−1) 𝑦′= 5𝑥4cos 𝑥5 b.𝑦 = sin5𝑥 (1 mark)
𝑦′= (sin5𝑥)′ 𝑦′= 5 sin5−1𝑥(𝑠𝑖𝑛𝑥)′ 𝑦′= 5 sin4𝑐𝑜𝑠𝑥 c.𝑦 = cos(4𝑥 + 9 ) (1 mark)
𝑦′= (cos (4𝑥 + 9)′ 𝑦′= − sin (4𝑥 + 9)(4𝑥 + 9) 𝑦′= − 𝑠𝑛(4𝑥 + 9 )(4) 𝑦′= −4 sin(4𝑥 + 9) ∴ The derivative of 𝑦=𝑠𝑖𝑛𝑥5 is 𝑦′=5𝑥4cos𝑥5 ∴ The derivative of 𝑦=sin5𝑥 is 𝑦′=5sin4𝑐𝑜𝑠𝑥 ∴ The derivative of 𝑦=cos(4𝑥+9) is 𝑦′=−4sin(4𝑥+9) Unit # 4 – Extensions d.𝑦 = cos2(𝑥2+ 3) (2 marks)
𝑦′= (cos2(𝑥2+ 3))′ 𝑦′= 2 cos(𝑥2+ 3)(cos(𝑥2+ 3) 𝑦′= 2cos (𝑥2+ 3)(−sin (𝑥2+ 3)(𝑥2+ 3) 𝑦′= 2co s (𝑥2+ 3)(−sin (𝑥2+ 3)(𝑥2+ 3)′ 𝑦′= −2cos (𝑥2+ 3)𝑠 in (𝑥2+ 3)(2𝑥) 𝑦′= −4𝑥 cos (𝑥2+ 3)𝑠 in (𝑥2+ 3) e.𝑦 = 𝑥2sin(𝑥2+ 5) (3 marks)
𝑦′ = 𝑥2sin(𝑥2+ 5) ′ 𝑦′= 𝑥2(sin(𝑥2+ 5)′+ sin(𝑥2+ 5)(𝑥2)′ 𝑦′= 𝑥2cos (𝑥2+ 5)(𝑥 62+ 5)′ + sin(𝑥2+ 5)(2𝑥2−1) 𝑦′= 𝑥2cos(𝑥2+ 5) (2𝑥)+ sin (𝑥2+ 5)(2𝑥) 𝑦′= 2𝑥3cos(𝑥2+ 5)+ 2𝑥 sin(𝑥2+ 5) ∴ The derivative of 𝑦=cos2(𝑥2+3) is 𝑦′=−4𝑥 cos(𝑥2+3)𝑠in(𝑥2+3) ∴ The derivative of 𝑦=𝑥2sin(𝑥2+5) is 𝑦′=2𝑥3cos(𝑥2+5)+2𝑥sin(𝑥2+5)
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