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Assignment No. 02 (Spring 2019) (Homework)
Mansour Mohammed Saeed Alkaabi
MATH105_20357_Spring_2019, section Calculus I (02), Spring 2019
Instructor: WebAssign Vendor
Current Score : 54 / 61 Due : Friday, March 1, 2019 11:59 PM +04
1. 1/1 points | Previous AnswersSCalcET8M 3.2.011.MI.
Differentiate.
34y2+21y4+5
2. 1/1 points | Previous AnswersSCalcET8M 3.2.031.
Find an equation of the tangent line to the given curve at the specified point.
WebAssign
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F(y) = − (y + 5y3)
1
y2
7
y4
y = , (1, 0)x2 − 1
x2 + x + 1
y =
23x−23
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3. 2/2 points | Previous AnswersSCalcET8M 3.2.036.
(a) The curve is called a serpentine. Find an equation of the tangent line to this curve at the point
y =
−450x+2750
(b) Illustrate part (a) by graphing the curve and tangent line on the same screen.
4. 3/3 points | Previous AnswersSCalcET8M 3.2.043.
Suppose that Find the following values.
(a)
-18
(b)
-54/81
(c)
6
y = x/(1 + x2) (3, 0.30).
f(5) = 3, f '(5) = 4, g(5) = −9, and g'(5) = 6.
(fg)'(5)
(f/g)'(5)
(g/f)'(5)
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5. 4/4 points | Previous AnswersSCalcET8M 3.2.052.
If f is a differentiable function, find an expression for the derivative of each of the following functions.
(a)
(b)
(c)
(d)
y = x7f(x)
y' = x6f '(x) − 6x7f(x)
y' = x6f '(x) + 6x7f(x)
y' = x7f(x) + 7x6f '(x)
y' = x7f '(x) + 7x6f(x)
y' = x7f(x) − 7x6f '(x)
y = f(x)
x8
y' = xf(x) + 8f '(x)
x8
y' = xf '(x) + 7f(x)
x8
y' = xf '(x) − 7f(x)
x9
y' = xf(x) − 8f '(x)
x7
y' = xf '(x) − 8f(x)
x9
y = x3
f(x)
y' = f '(x)(3x2) − x3f(x)
[f(x)]2
y' = f '(x)(2x3) + x2f(x)
[f(x)]2
y' = f(x)(2x3) + x3f '(x)
[f(x)]2
y' = f '(x)(2x2) − x3f '(x)
[f(x)]2
y' = f(x)(3x2) − x3f '(x)
[f(x)]2
y = 3 + xf(x)
x
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6. 1/1 points | Previous AnswersSCalcET8M 3.3.022.
Find an equation of the tangent line to the curve at the given point.
y =
5x+5
7. 3/3 points | Previous AnswersSCalcET8M 3.3.031.
(a) Use the Quotient Rule to differentiate the function
(b) Simplify the expression for by writing it in terms of sin(x) and cos(x), and then find
(c) Are your answers to parts (a) and (b) equivalent?
y' = xf '(x) − 2x2f '(x) + 3
2x3/2
y' = xf(x) − 3x2f '(x) + 3
2x3/2
y' = xf(x) + 3x2f(x) − 2
2x3/2
y' = xf '(x) + 3x2f(x) − 2
2x3/2
y' = xf(x) + 2x2f '(x) − 3
2x3/2
y = 5ex cos(x), (0, 5)
f(x) = .
tan(x) − 1
sec(x)
f '(x) =
sec3(x)−tan2(x)sec(x)+tan(x)sec(x)sec2(x)
f(x) f '(x).
f '(x) =
cos(x)+sin(x)
Yes
No
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