MTH131 Applied Calculus – Spring 2016
Lab 3 – Derivatives
SOLUTIONS
1. (a) f 0 (x) = 4x3
(b) g 0 (x) = 32 x2
(c) g 0 (w) = 2w−2/3
6
(d) h0 (x) = − 3
x
(e) f 0 (x) = 8x − 3
(f) f 0 (r) = 2πr
1
1
(g) f 0 (x) = √ − 2
2 x x
2. In all there problems, you can stop simplifying when you get to lines marked with a (*).
(a) f (x) = x3 (x2 + 1): Using the product rule
f 0 (x) = 3x2 (x2 + 1) + x3 (2x)
= 3x2 (x2 + 1) + 2x4 (*)
= 3x4 + 3x2 + 2x4
= 5x4 + 3x2
(b) g(x) = (x3 − 1)(x3 + 1): Using the product rule
g 0 (x) = 3x2 (x3 + 1) + (x3 − 1)(3x2 ) (*)
= 3x5 + 3x2 + 3x5 − 3x2
= 6x5
(c) g(t) = (2t − 1)(1 − t2 ): Using the product rule
g 0 (t) = 2(1 − t2 ) + (2t − 1)(−2t)
= 2(1 − t2 ) − (2t − 1)(2t) (*)
= 2 − 2t2 − 4t2 + 2t
= −6t2 + 2t + 2
(d) f (x) = x3 (x2 − 4x + 3): Using the product rule
f 0 (x) = 3x2 (x2 − 4x + 3) + x3 (2x − 4) (*)
= 3x4 − 12x3 + 9x2 + 2x4 − 4x3
= 5t4 − 16x3 + 9x2
√
√
(e) f (x) = (x + 6 x)(x − 2 x + 1) = (x + 6x1/2 )(x − 2x1/2 + 1): Using the product rule
f 0 (x) = (1 + 6(1/2)x−1/2 )(x − 2x1/2 + 1) + (x + 6x1/2 )(1 − 2(1/2)x−1/2 )
= (1 + 3x−1/2 )(x − 2x1/2 + 1) + (x + 6x1/2 )(1 − x−1/2 ) (*)
= (x + x1/2 − 5 + 3x−1/2 ) + (x + 5x1/2 − 6)
= 2x + 6x1/2 − 11 + 3x−1/2
(f) h(x) =
x5 − 1
: Using the quotient rule:
x2
h0 (x) =
=
=
=
(g) g(t) =
x2 (5x4 ) − (x5 − 1)(2x)
(x2 )2
6
5
5x − (x − 1)(2x)
(*)
x4
5x6 − 2x6 + 2x
x4
6
3x + 2x
x4
t2 + 1
: Using the quotient rule:
t2 − 1
(t2 − 1)(2t) − (t2 + 1)(2t)
(*)
(t2 − 1)2
2t3 − 2t − 2t3 − 2t
=
(t2 − 1)2
4t
= − 2
(t − 1)2
g 0 (t) =
(h) f (x) =
x5 + x3 + x
: Using the quotient rule:
x3 + x
f 0 (x) =
=
=
(x3 + x)(5x4 + 3x2 + 1) − (x5 + x3 + x)(3x2 + 1)
(*)
(x3 + x)2
(5x7 + 8x5 + 4x3 + x) − (3x7 + 4x5 + 4x3 + x)
(x3 + x)2
2x7 + 4x5
(x3 + x)2
√
x+1
(i) f (x) = √
: Using the quotient rule:
x−1
f 0 (x) =
=
(x1/2 − 1)( 12 x−1/2 ) − (x1/2 + 1)( 12 x−1/2 )
(*)
(x1/2 − 1)2
(1/2 − 12 x−1/2 ) − (1/2 + 12 x−1/2 )
√
( x − 1)2
x−1/2
= − √
( x − 1)2
1
= −√ √
x( x − 1)2
(x5 + 1)(x3 + 2)
: For this problem, you must use both the product and quotient
x+1
rules. You can do them in either order. The answer below uses the quotient rule first,
then the product rule.
(j) f (x) =
f 0 (x) =
=
=
=
=
=
d
d
(x + 1) dx
[(x5 + 1)(x3 + 2)] − (x5 + 1)(x3 + 2) dx
[x + 1]
2
(x + 1)
4
3
(x + 1)[(5x )(x + 2) + (x5 + 1)(3x2 )] − (x5 + 1)(x3 + 2)
(*)
(x + 1)2
(x + 1)[(5x7 + 10x4 ) + (3x7 + 3x2 )] − (x8 + 2x5 + x3 + 2)
(x + 1)2
7
4
2
(x + 1)(8x + 10x + 3x ) − (x8 + 2x5 + x3 + 2)
(x + 1)2
(8x8 + 8x7 + 10x5 + 10x4 + 3x3 + 3x2 ) − (x8 + 2x5 + x3 + 2)
(x + 1)2
7x8 + 8x7 + 8x5 + 10x4 + 2x3 + 3x2 − 2
(x + 1)2
3. The derivative of f (x) =
x+1
is
x−1
(x − 1)(1) − (x + 1)(1)
(x − 1)2
2
= −
.
(x − 1)2
f 0 (x) =
We want to know where this function is = −2:
2
(x − 1)2
1
(x − 1)2
(x − 1)2
2
x − 2x + 1
x2 − 2x
x(x − 2)
x
−
= −2
divide by −2
= 1
invert
=
=
=
=
=
expand
1
1
0
0
0, 2
factor, and solve for x
At x = 0 the function will have the value f (0) = (0 + 1)/(0 − 1) = −1. Given that the slope
= −2 at this point, we use the point-slope formula for a line to get the equation of the tangent
line:
(y − (−1)) = −2(x − 0),
y + 1 = −2x,
y = −2x − 1.
At x = 2 the function will have the value f (2) = (2 + 1)/(2 − 1) = 3. Given that the slope
= −2 at this point, we use the point-slope formula for a line to get the equation of the tangent
line:
(y − 3) = −2(x − 2),
y − 3 = −2x + 4,
y = −2x + 7.
4. (a) The rate of change of temperature is given by the derivative f 0 (t) = 8t1/3 . We need to
know when this is equal to 16 degress per minute:
8t1/3 = 16,
t1/3 = 2,
t = 8.
So after 8 minutes, the temperature is changing at 16 degress per minute.
(b) It takes 8 minutes to reach the point when the temperature is changing at 16 degrees/minute. At that point, the actual temperature is f (8) = 24 + 6(8)4/3 = 120.
If the metal is now cooled at 10 degrees per minute, it takes 12 more minutes to get to
0 degress Celcius. Therefore, the entire process takes 8+12 = 20 minutes.
5. (a) f (x) = x3 (x2 + 1): Simplifying f (x) first
f (x) = x5 + x3 ,
f 0 (x) = 5x4 + 3x2
(b) g(x) = (x3 − 1)(x3 + 1): Simplifying g(x) first
g(x) = x6 − 1 using rule (a + b)(a − b) = a2 − b2 ,
g 0 (x) = 6x5
(c) g(t) = (2t − 1)(1 − t2 ): Simplifying g(t) first
g(t) = −2t3 + t2 + 2t − 1,
g 0 (t) = −6t2 + 2t + 2
(d) f (x) = x3 (x2 − 4x + 3): Simplifying f (x) first
f (x) = x5 − 4x4 + 3x3
f 0 (x) = 5x4 − 16x3 + 9x2