Math 2215 - Calculus 1
Exam #6 - 2016.11.30
Solutions
d
1
sin−1 (x) = √
dx
1 − x2
d
1
cot−1 (x) = −
dx
1 + x2
d
1
sinh−1 (x) = √
dx
1 + x2
d
1
coth−1 (x) =
dx
1 − x2
Z
1. Compute the following integral:
d
1
cos−1 (x) = − √
dx
1 − x2
d
1
√
sec−1 (x) =
dx
|x| x2 − 1
d
1
cosh−1 (x) = √
2
dx
x −1
d
1
−1
sech (x) = − √
dx
x 1 − x2
d
1
tan−1 (x) =
dx
1 + x2
d
1
csc−1 (x) = − √
dx
|x| x2 − 1
d
1
tanh−1 (x) =
dx
1 − x2
d
1
csch−1 (x) = − √
dx
|x| x2 + 1
sin(ln(2x))
dx
x
1
We perform a u-substitution: u = ln(2x), and du = dx.
x
Z
Z
sin(ln(2x))
dx = sin(u) du
x
= − cos(u) + C
= − cos(ln(2x)) + C
2. Compute the following derivative:
d
sin(x2 )cos(x)+x
dx
First we set y = sin(x2 )cos(x)+x , and thus ln(y) = (cos(x) + x) ln(sin(x2 )), We can now implicitly differentiate:
d
d ln(y) =
(cos(x) + x) ln(sin(x2 ))
dx
dx
y0
cos(x2 )2x
= (− sin(x) + 1) ln(sin(x2 )) + (cos(x) + x)
y
sin(x2 )
Solving for y 0 gives
cos(x2 )2x
y 0 = (− sin(x) + 1) ln(sin(x2 )) + (cos(x) + x)
y
sin(x2 )
cos(x2 )2x
= (− sin(x) + 1) ln(sin(x2 )) + (cos(x) + x)
sin(x2 )cos(x)+x
sin(x2 )
3. Derive the formula for
d
tan−1 (x) by the method of implicit differentiation.
dx
Setting y = tan−1 (x) gives tan(y) = x. We then implicitly differentiate:
d
d
tan(y) =
x
dx
dx
2
0
sec (y)y = 1
Solving for y 0 gives
y0 =
1
sec2 (y)
= cos2 (y)
= cos2 tan−1 (x)
1
2
Setting θ = tan−1 (x), and thus tan(θ) = x. Drawing a triangle gives opposite side x, adjacent side 1 and hypotenuse
√
1
1 + x2 . Thus cos(θ) = √
and we finally end up with
1 + x2
d
1
tan−1 (x) =
dx
1 + x2
4. Verify algebraically that cosh2 (x) − sinh2 (x) = 1.
First, we rewrite sinh(x) and cosh(x) in terms of their exponentials:
x
2 x
2
e + e−x
e − e−x
−
=1
2
2
And then we multiply it all out:
e2x + 2 + e−2x
e2x − 2 + e−2x
2+2
−
=
=1
4
4
4
Z
5. Compute the following integral: 3x tanh(x2 ) dx
3
We perform a u-substitution: u = x2 , and du = 2x dx, and thus 3x dx = du.
2
Z
Z
3
2
3x tanh(x ) dx =
tanh(u) du
2
Z
3 sinh(u)
=
du
2 cosh(u)
3
= ln (|cosh(u)|) + C
2
3
= ln cosh(x2 ) + C
2
3
= ln cosh(x2 ) + C
2
On the last line above, since cosh(z) ≥ 1 for all z, the absolute value can be removed.
(5x − 1)2 (3x + 2)5
d
ln
.
6. Compute the following derivative:
dx
(4x − 7)6 (3x − 2)7
First we simplify our function using properties of logarithms:
(5x − 1)2 (3x + 2)5
ln
= 2 ln(5x − 1) + 5 ln(3x + 2) − 6 ln(4x − 7) − 7 ln(3x − 2)
(4x − 7)6 (3x − 2)7
What we now have is easy to differentiate:
(5x − 1)2 (3x + 2)5
d
d
ln
[2 ln(5x − 1) + 5 ln(3x + 2) − 6 ln(4x − 7) − 7 ln(3x − 2)]
=
dx
(4x − 7)6 (3x − 2)7
dx
5
3
4
3
=2
+5
−6
−7
5x − 1
3x + 2
4x − 7
3x − 2
Z
9x
√
7. Compute the following integral:
dx
2
x 1 − x4
9
We perform a u-substitution: u = x2 , and du = 2x dx, and thus 9x dx = du.
2
Z
Z
9x
9
1
√
√
dx =
du
2 u 1 − u2
x2 x4 − 1
9
= − sech−1 (u) + C
2
9
= − sech−1 (x2 ) + C
2
3
d
log7 23x − 45x + 1 .
dx
d
1 d
log7 23x − 45x + 1 =
ln 23x − 45x + 1
dx
ln(7) dx
d
23x − 45x + 1
1 dx
=
ln(7) 23x − 45x + 1
1 ln(2)23x · 3 − ln(4)45x · 5
=
ln(7)
23x − 45x + 1
3
2
9. If f (x) = x + 3x + 4x − 1, verify that f (x) is invertible by showing that f 0 (x) > 0 for all x. Then compute
d −1
f (x) at x = 7.
dx
8. Compute the following derivative:
First, f 0 (x) = 3x2 + 6x + 4, and computing the discriminant gives d = 36 − 48 = −12 < 0, thus there are no roots
to the derivative and thus f 0 (x) > 0 which means the function is always increasing. Thus, f (x) is invertible.
To compute the derivative of the inverse at x = 7, we need to find f −1 (7). By plugging simple small numbers into
f (x), we see that f (1) = 7, thus f −1 (7) = 1. So now we have that
1
d −1 = 0 −1
f (x)
dx
f (f (7))
x=7
1
f 0 (1)
1
=
13
=