Math 1LS3 - Winter 2014
Tutorial 7
Spencer Hunt
February 25, 2015
Topics Covered
• Derivatives of different functions
– Exponential functions
– Logarithmic functions
– Trigonometric functions
– Inverse trigonometric functions
• Differentiation techniques
– Product Rule
– Quotient Rule
– Chain rule
– Implicit differentiation
• Equations of the tangent line
• Multiple derivatives
Pre-Test Questions
1. True or False
The 2nd derivative of the function f (x) = 3x is equal to 3x ln(9)
2. True or False
The derivative of the function g(x) = (2x3 − 8)ex is equal to (6x2 )(ex )
3. True or False
The derivative of the function h(x) = (3x2 − x + 4)3 is equal to 3(6x − 1)2 .
1
Tutorial Questions
1. Calculate the derivatives of the following functions
a) h(x) = cos(x3 − 3x + 4)
h0 (x) = − sin(x3 − 3x + 4)(x3 − 3x + 4)0 , Chain Rule
= − sin(x3 − 3x + 4)(3x2 − 3)
(1)
(2)
b) f (x) = e2x sin(x2 )
f 0 (x) = (e2x )0 (sin(x2 )) + (e2x )(sin(x2 ))0 , Product Rule
= (e2x (2))(sin(x2 )) + e2x (cos(x2 )(2x)), Chain Rule
= e2x (2 sin(x) + 2x cos(x2 ))
(3)
(4)
(5)
c) g(x) = ln(x) sec(x)
ln(x)
cos(x)
(ln(x))0 (cos(x)) − (ln(x))(cos(x))0
g 0 (x) =
, Quotient Rule
[cos(x)]2
1
cos(x) − ln(x)(− sin(x))
= x
[cos x]2
g(x) =
=
cos(x)+x ln(x) sin(x)
x
[cos(x)]2
=
cos(x) + x ln(x) sin(x)
x[cos(x)]2
(6)
(7)
(8)
(9)
(10)
1
2. Find the equation of the tangent line to the function f (x) = e3x − at x = 1.
x
In order to find the equation of the tangent line we need two things. We need a
point that the function and the tangent line both go through, which in this case
is (x, f (x)) and we need the slope of the line m = f 0 (x), for x = 1.
f 0 (x) = 3e3x +
1
x2
m = f 0 (1) = 3e3(1) +
= 3e3 + 1
2
1
(1)2
(11)
(12)
(13)
Now we will solve for the point at the tangent line meets the function.
f (1) = e3(1) −
1
1
= e3 − 1
(14)
(15)
so we go through the point P = (1, e3 −1). Putting this together in the point-slope
formula for the tangent line we have
y − y0 = m(x − x0 )
y − (e3 − 1) = (3e3 + 1)(x − 1)
y = (3e3 + 1)x − (3e2 + 1) + (e3 − 1)
y = (3x3 + 1)x − 2e3
(16)
(17)
(18)
(19)
1
1
, prove that (loga (x))0 =
.
x
x ln(a)
Recall that we can rewrite loga (x) using the change of base law of logarithms.
3. Using the fact that (ln(x))0 =
loga (x) =
ln(x)
ln(a)
(20)
From here, we can just take the derivative, noting that ln(a) is just a constant.
0
ln(x)
(ln(x))0
1
=
=
(21)
ln(a)
ln(a)
x ln(a)
which proves what we were trying to show.
K
4. Calculate the second derivatives of the following functions
a) f (x) = 2x
f 0 (x) =2x (ln(2))
f 00 (x) =2x (ln(2))(ln(2))
f 00 (x) =2x (ln(4))
(22)
(23)
(24)
b) g(x) = arctan(x)
1
= (1 + x2 )−1
1 + x2
g 00 (x) = −1(1 + x2 )−2 (2x)
−2x
=
(1 + x2 )2
g 0 (x) =
3
(25)
(26)
(27)
c) h(x) = arcsin(2x)
2
h0 (x) = √
= 2(1 − x2 )−1/2
2
1−x
00
h (x) = 2( −1
)(1 − x2 )−3/2 (−2x)
2
2x
=
(1 − x2 )3/2
(28)
(29)
(30)
Just a note on notation (ha!), the expression f (n) (x) is considered to be the n-th
derivative of the function f (x). This means that you take the derivative n times.
Generally, we use this notation when we take the third derivative or higher.
5. Calculate the following derivatives of the functions
a) Calculate f (4) (x) of the function f (x) = e2x .
The trick to these questions is to spot the pattern when we take the derivatives
f 0 (x) = 2e2x
(31)
f 00 (x) = 2(2)e2x = 22 e2x
(32)
and then the second derivative
So you will notice that e2x stays exactly the same, but for every derivative
we take we multiply by 2. Which means when we take the fourth derivative
f (4) (x) = 24 e2x
(33)
b) Calculate the first four derivatives of the function g(x) = sin(x).
f 0 (x) = (sin(x))0 = cos(x)
f 00 (x) = (cos(x))0 = − sin(x)
f 000 (x) = (− sin(x))0 = − cos(x)
f (4) (x) = (− cos(x))0 = sin(x)
(34)
(35)
(36)
(37)
c) Calculate the derivative g (1001) (x) of the function g(x) = sin(x).
Notice that in the above question, we showed that after four derivatives,
we obtain the exact same function again. Which means that for every four
multiples of derivative we obtain the same function sin(x). We can therefore
simplify this significantly.
g (1001) (x) = g (4(250)+1) (x) = g 0 (x) = cos(x)
4
(38)
6. Using implicit differentiation find
dy
from the expression
dx
3y 2 − x sin(y) = 9x2 − x
(39)
Implicit differentiation is an application of chain rule. We assume that we have a
function y(x), and an expression f (y(x)), so in order to solve for the derivative,
dy
each
we can take the derivative of f in terms of y making sure to multiply by dx
dy
time (just like in chain rule) and then solve for dx .
3(2)y
dy
dy
− (1(sin(y)) + x cos(y) ) = 18x − 1
dx
dx
dy
(6y − x cos(y)) = 18x − 1 + sin(x)
dx
dy
18x − 1 + sin(x)
=
dx
6y − x cos(y)
5
(40)
(41)
(42)