MCV4U d3-D.E.L.F. Assignment
Answer all questions with full solutions. Make sure your work is legible, even after you have
scanned it, and submit it as 0 single file. Note that derivatives must be fully simplified for full
marks.
1. Find the derivatives of each of these functions:
a.y= ln (4x^5+1)
b.y=(4x^5)/e^3x
c.y=5^(2x+6) ( ln (5x+7))
d.y=(4x^2)/((e^6x+x^4))
2. Use the process of implicit differentiation to find dy/dx given that xe^y-y^3 e^x=0 3. Sketch the graph of the function y=x^2 e^X
4. Use logarithmic differentiation to find the derivative of the function:
5. The number of malaria parasites(p)in the blood h hours aftera patient takes medication is modelled by the function p=200/(0.9+0.1e^2h )
a. What is the initial number of parasites in the blood?
b. How longdid it take before the number of parasites had been halved?
c. After how many hours was the number of parasites decreasing the fastest?
d. Sketch the function.
e. Give a possible explanation for the shape of the curve
D.E.L.F. Assignment - Virtual High School (VHS) - MCV4U
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MCV4U d3, Calculus and Vectors,12, Virtual High School (VHS)
Unit Assignment: DELF
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2. Use the process of implicit
differentiation to find
𝑑𝑦
𝑑𝑥
given that 𝑥𝑒 𝑦 − 𝑦3 𝑒 𝑥 = 0
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3. Sketch the graph of the
function 𝑦 = 𝑥2 𝑒 𝑋
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4. Use logarithmic
differentiation to find the
derivative of the function:
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5. The number of malaria parasites (p) in the
blood h hours after a patient takes
medication is modelled by the function
𝑝 =
200
0.9 + 0.1𝑒2ℎ
a. What is the initial number of parasites in
the blood?
b. How long did it take before the number of
parasites had been halved?
c. After how many hours was the number of
parasites decreasing the fastest?
d. Sketch the function.
e. Give a possible explanation for the shape
of the curve.
![Assignment Help | 100% Plagiarism Free | Success Assured | Email Now to get quote – admin@panhelp.com
1. Find the derivatives of each of
these functions:
𝑎. 𝑦 = ln (4𝑥5 + 1)
𝑏. 𝑦 =
4𝑥5
𝑒3𝑥
𝑐. 𝑦 = 52𝑥+6( ln (5𝑥 + 7))
𝑑. 𝑦 =
4𝑥2
(𝑒6𝑥+𝑥4)
1. A) 𝑦 = ln 4𝑥5
+ 1
𝑑𝑦
𝑑𝑥
=
1
4𝑥5+1
20𝑥4
𝑑𝑦
𝑑𝑥
=
20𝑥4
4𝑥5+1
B) Y =
4𝑥5
𝑒3𝑥
𝑑𝑦
𝑑𝑥
=
20𝑥4 𝑒3𝑥 − 4𝑥5 𝑒3𝑥 3
𝑒3𝑥 2
𝑑𝑦
𝑑𝑥
=
20𝑥4−12𝑥5
𝑒3𝑥 =
4𝑥4(5−3𝑥)
𝑒3𝑥
C) 𝑦 = 52𝑥+6
ln(5𝑥 + 7)
𝑑𝑦
𝑑𝑥
= 52𝑥+6
ln 5𝑥 + 7 (ln 5)2+52𝑥+6 1
5𝑥+7
(5)
𝑑𝑦
𝑑𝑥
= 52𝑥+6
[ln 5𝑥 + 7 (ln 5)2+
1
5𝑥+7
(5)]
D) 𝑦 =
4𝑥2
𝑒6𝑥+𝑥4
𝑑𝑦
𝑑𝑥
=
8𝑥 𝑒6𝑥+𝑥4 −(4𝑥2)(𝑒6𝑥(6)+4𝑥3)
𝑒6𝑥+𝑥4 2
𝑑𝑦
𝑑𝑥
=
8𝑥 [ 𝑒6𝑥+𝑥4 −(𝑥)(𝑒6𝑥(3)+2𝑥3)
𝑒6𝑥+𝑥4 2](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)