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Math 113 HW #6 Solutions
- Exercise 3.1.22. Differentiate the function
y =
x(x − 1)
Answer: Re-write the function as
y = x
x −
x = x^3 /^2 − x^1 /^2.
Then, using the power rule,
y′^ =^3 2
x^1 /^2 − 1 2
x−^1 /^2 =^3 2
x − 1 2
x
- Exercise 3.1.34. Find an equation of the tangent line to the curve
y = x^4 + 2x^2 − x
at the point (1, 2). Answer: Using the power rule, y′^ = 4x^3 + 4x − 1 , so, at the point (1, 2), y′^ = 4(1)^3 + 4(1) − 1 = 7. Therefore the slope of the tangent line is 7. Since the tangent line passes through the point (1, 2), we can use the point-slope formula:
y − 2 = 7(x − 1),
so, adding 2 to both sides, we have that
y = 7x − 5.
- Exercise 3.2.24. Differentiate f (x) =
1 − xex x + ex^. Answer: Using the quotient and product rules,
f ′(x) = (x^ +^ e
x)(−ex (^) − xex) − (1 − xex)(1 + ex) (x + ex)^2
= −xe
x (^) − x (^2) ex (^) − e 2 x (^) − xe 2 x (^) − (1 + ex (^) − xex (^) − xe 2 x) (x + ex)^2
=
− 1 − (1 + x^2 + ex)ex (x + ex)^2.
- Exercise 3.2.28. Let f (x) = x^5 /^2 ex and find f ′(x) and f ′′(x). Answer: Using the product rule,
f ′(x) =
2 x
3 / 2
ex^ + x^5 /^2 ex^ =
2 x
3 / (^2) ex (^) + x 5 / (^2) ex. (1)
Taking the derivative of the first term yields 5 2
2 x
1 / (^2) ex (^) + x 3 / (^2) ex
=^154 x^1 /^2 ex^ +^52 x^3 /^2 ex,
whereas the second term in (1) is just a copy of f , so its derivative is the same as f ′(x). Therefore,
f ′′(x) =
4 x
1 / (^2) ex (^) +^5 2 x
3 / (^2) ex
=
4 x
1 / (^2) ex (^) +^5 2 x
3 / (^2) ex
2 x
3 / (^2) ex (^) + x 5 / (^2) ex
4 x
1 / (^2) ex (^) + 5x 3 / (^2) ex (^) + x 5 / (^2) ex.
- Exercise 3.2.34. Find the equations of the tangent line and normal line to the curve
y =
x x + 1 at the point (4, 0 .4). Answer: Writing
x as x^1 /^2 and using the quotient rule,
y′^ =
(x + 1)
2 x−^1 /^2
− x^1 /^2 (1) (x + 1)^2
=
1 2 x
2 x
− 1 / (^2) − x 1 / 2 (x + 1)^2
=^1 2
x−^1 /^2 − x^1 /^2 (x + 1)^2
=^1 2
√^1 x −
x (x + 1)^2 Therefore, we can plug in x = 4 to get the slope of the tangent line at (4, 0 .4):
1 2
√^1 4 −
(4 + 1)^2
=^1
1 2 −^2 25
Therefore, we can use the point-slope formula to get the equation of the tangent line:
y − 2 5
(x − 4),
- Exercise 3.3.34. Find the points on the curve y = (cos x)/(2 + sin x) at which the tangent is horizontal. Answer: The tangent being horizontal means that the slope of the tangent line is zero. In other words, we’re looking for those points where the derivative of the function is zero. Using the quotient rule, the derivative is
y′^ = (2 + sin^ x)(−^ sin^ x)^ −^ cos^ x(cos^ x) (2 + sin x)^2
= −2 sin^ x^ −^ sin
(^2) x − cos (^2) x (2 + sin x)^2
Using the fact that sin^2 x + cos^2 x = 1, this simplifies to
y′^ = −2 sin^ x^ −^1 (2 + sin x)^2
which is zero if and only if the numerator is zero. Therefore, we want to find the values of x for which 0 = −2 sin x − 1 or, equivalently,
sin x = − 12.
This equality holds when
x =... , −^5 π 6
, −π 6
, 7 π 6
, 11 π 6
, 19 π 6
, 23 π 6
,... , (12n^ −^ 5)π 6
, (12n^ −^ 1)π 6
- Exercise 3.4.12. Find the derivative of
f (t) = 3
1 + tan t.
Answer: Re-writing as f (t) = (1 + tan t)^1 /^3 = g ◦ h(t), where g(u) = u^1 /^3 and h(t) = 1 + tan t, then the Chain Rule tells us that f ′(t) = g′(h(t)) · h′(t).
Now, g′(u) = 13 u−^2 /^3 and h′(t) = sec^2 t, so
f ′(t) = g′(h(t)) · h′(t) =
3 (h(t))
− 2 / (^3) · sec (^2) t =^1 3 (1 + tan^ t) sec
(^2) t = sec^2 t 3(1 + tan t)^2 /^3
- Exercise 3.4.22. Find the derivative of the function
y = e−^5 x^ cos 3x.
Answer: By the Chain Rule, the derivative of h 1 (x) = e−^5 x^ = f 1 circg 1 (x) where f 1 (u) = eu and g 1 (x) = − 5 x is
h′ 1 (x) = f 1 ′(g 1 (x)) · g′ 1 (x) = eg^1 (x)(−5) = − 5 e−^5 x.
Also using the Chain Rule, the derivative of h 2 (x) = cos 3x = f 2 ◦ g 2 (x) where f 2 (u) = cos u and g 2 (x) = 3x is
h′ 2 (x) = f 2 ′(g 2 (x)) · g 2 ′(x) = − sin(g 2 (x)) · (3) = −3 sin 3x.
Therefore, using these facts and the product rule,
y′^ = − 5 e−^5 x^ cos 3x + e−^5 x(−3 sin 3x) = − 5 e−^5 x^ cos 3x − 3 e−^5 x^ sin 3x.
- Exercise 3.4.34. Find the derivative of the function
y = x sin^1 x
Answer: By the Chain Rule, the derivative of h(x) = sin (^1) x = f ◦ g(x) where f (u) = sin u and g(x) = 1/x = x−^1 is
h′(x) = f ′(g(x)) · g′(x) = cos(g(x)) · (−x−^2 ) = −
cos (^) x^1 x^2
Therefore, using the product rule,
y′^ = 1 · sin^1 x − x ·
cos (^1) x x^2 , so y′^ = sin
x −^
x cos
x.
