Beware!!!
d 2x
e 6= e2x .
dx
The function e2x is going up twice as fast as ex is, so its derivative should
be twice as steep. To make this precise, we’ll use (soon!) the chain rule.
3.3
Lecture 8: Derivatives of Trigonometric Functions, plus product rule,
quotient rule
We are building a set of functions of which we can compute the derivatives (and hopefully also anti-derivatives).
Since both the derivative and the integral are linear operators, we can also deduce the derivatives and antiderivatives of sums and scalar multiples of these functions.
So far we have
• Power rule: if f (x) = xn where n ∈ R then f ′ (x) = nxn−1 . This works for all n ∈ R although right now we
have only the tools to prove it for certain values of n.
• Exponential: if f (x) = ex then f ′ (x) = ex , where e is defined to be the number such that the derivative of
f (x) = ex at x = 0 is exactly 1.
−6/5 + ex .
For example, f (x) = 2x4/5 + ex implies f ′ (x) = 85 x−1/5 + ex and f ′′ (x) = −8
25 x
These also imply corresponding rules for some anti-derivatives:
R
1
xn+1 + c.
• Anti-power rule: if f (x) = xn and n 6= −1 then f (x) dx = n+1
R
• Exponential: if f (x) = ex then ex dx = ex + c.
R2
Example 3.14. Find 0 ex+3 dx.
We use the FTC.
Z 2
Z 2
x+3
e3 ex dx
e dx =
0
0
Z 2
3
=e
ex dx
0
= e [ex ]20
3
= e3 (e2 − e0 )
= e3 (e2 − 1)
= e5 − e3
Example 3.15. If f (x) = ex + x4 , then find f ′′ (x).
We have f ′ (x) = ex + 4x3 so f ′′ (x) = ex + 12x2 .
Let’s tackle the trigonometric functions today.
3.3.1
Derivatives of sine and cosine
We sketch the graph of the sine function, and underneath, we roughly sketch the graph of its derivative, noting
where it is positive, zero and negative. We do the same for the cosine function. In both cases, we realize that the
derivative is again a periodic function of period 2π. Also, we roughly guess at the maximum slope of the sine and
cosine curves (which corresponds to the amplitude of their derivative function) with the slope of the secant line
from (0, 0) on y = sin(x) to (π/6, 1/2) and get 3/pi ∼ 0.95 — because we are measuring angles in radians!!
If we used degrees, we’d get slope around 0.017. USE RADIANS IN ALL OF CALCULUS.
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Now we tackle the derivative from the definition.
For f (x) = sin(x), the difference quotient at x is:
sin(x + h) − sin(x)
sin(x) cos(h) + sin(h) cos(x) − sin(x)
=
h
h
sin(h)
cos(h) − 1
+ cos(x)
= sin(x)
h
h
Which we cannot simplify any further. Similarly, if we try f (x) = cos(x), the difference quotient is:
cos(x + h) − cos(x)
cos(x) cos(h) − sin(x) sin(h) − cos(x)
=
h
h
sin(h)
cos(h) − 1
− sin(x)
= cos(x)
h
h
which is similar but definitely not the same as for sin(x). In any case, it isn’t any simpler.
So our problem is we need to figure out
cos(h) − 1
h→0
h
and
lim
sin(h)
h→0
h
lim
if they exist.
We can do this geometrically:
For y = sin(x), we know that h is the arc length on the unit circle to the point with coordinates (cos(h), sin(h);
drawing a picture we see that sin(h) ≤ h, for any h > 0, so
sin(h)
≤ 1.
h
Now consider the triangle with base length 1 and angle h; its height is tan(h). We have tan(h) ≥ h which gives,
for cos(h) ≥ 0,
sin(h)
.
cos(h) ≤
h
Therefore the fraction
sin(h)
h
is squeezed between cos(h) and 1 and so as h → 0 and cos(h) → 1, we have no choice but to conclude that
lim
h→0
sin(h)
= 1.
h
For the other limit: we can use the identity cos(θ) = 1 − 2 sin2 (θ) or −2 sin2 (θ) = cos(2θ) − 1 applied to h = 2θ
to get
1 − cos(h)
sin2 (h/2)
sin(h/2)
= −2
= − sin(h/2)
h
h
h/2
and as h → 0, we see that sin(h/2) → 0 and
sin(h/2)
h/2
→ 1 so the product goes to 0. Hence
cos(h) − 1
= 0.
h→0
h
lim
Returning to the original question, we conclude that
d
sin(x + h) − sin(x)
sin(x) = lim
h→0
dx
h
cos(h) − 1
sin(h)
+ cos(x)
= lim sin(x)
h→0
h
h
= cos(x)
40
and
d
cos(x + h) − cos(x)
cos(x) = lim
h→0
dx
h
cos(h) − 1
sin(h)
= lim cos(x)
− sin(x)
h→0
h
h
= − sin(x)
This gives us a new pair of rules:
d
d
sin(x) = cos(x)
and
cos(x) = − sin(x)
dx
dx
We can remember which one gives a negative by remembering that cos(x) has negative slope for 0 < x < π/2,
whereas sin(x) is a positive function on that interval.
Example 3.16. Find the first 4 derivatives of f (x) = sin(x).
f ′ (x) = cos(x), f ′′ (x) = − sin(x), f (3) (x) = − cos(x), f (4) (x) = sin(x). Amazingly: after 4 iterations we get
back to our original function.3
On the integral side, we have the rules
Z
Z
sin(x) dx = − cos(x) + c
and
cos(x) dx = sin(x) + c
Notice that the integral of sin(x) is − cos(x), because the derivative of cos(x) is − sin(x).
R π/2
Example 3.17. Find 0 sin(x) dx. From the graph, we know this is positive.
π/2
R π/2
sin(x) dx = − cos(x)
= (− cos(π/2) − (− cos(0)) = 0 + 1 = 1.
0
0
(If we’d accidentally forgotten the minus sign, our answer would have come out to −1, which is clearly false.)
What about the derivatives of the the other four trigonometric functions? Well, they are all quotients of sine
and cosine; so let’s quickly recall the product and quotient rules for derivatives.
3.3.2
Product rule
Theorem 3.18. Let f and g be differentiable functions on a common domain A. Then the product function f g
is also differentiable, and its derivative is given by
d
(f (x)g(x)) = f (x)g ′ (x) + f ′ (x)g(x).
dx
Notice that the units make sense: if f and g are in m and t in s, then the units on the derivative and each of
the terms are m2 /s.
Proof. We work on the difference quotient and apply a brilliant trick:
f (x + h)g(x + h) − f (x + h)g(x) + f (x + h)g(x) − f (x)g(x)
f (x + h)g(x + h) − f (x)g(x)
= lim
h→0
h→0
h
h
g(x + h) − g(x) f (x + h) − f (x)
= lim f (x + h)
+
g(x)
h→0
h
h
= f (x)g ′ (x) + f ′ (x)g(x)
lim
since upon taking the limit as h → 0, we see that f (x + h) → f (x), and the difference quotients go to the
derivatives. So the limit exists and f g is differentiable with the stated derivative.
3
If these suggests to you that there must be √
some relationship between sine, cosine and the exponential function, then you’re
absolutely right: eix = cos(x) + i sin(x) where i = −1.
41
Example 3.19.
d 2 x
(x e ) = x2
dx
d x
e
dx
+
d 2 x
x e = x2 ex + 2xex = ex (x2 + 2x)
dx
The rule extends to products of 3 and more terms, via
(f gh)′ (x) = f (x)(gh)′ (x) + f ′ (x)(gh)(x)
= f (x)(g(x)h′ (x) + g ′ (x)h(x)) + f ′ (x)g(x)h(x)
= f (x)g(x)h′ (x) + f (x)g ′ (x)h(x) + f ′ (x)g(x)h(x)
Example 3.20.
d 1/3
d 4/3
(x + x1/3 )ex =
x (x + 1)ex
dx
dx
1
= x1/3 (x + 1)ex + x1/3 ex + x−2/3 (x + 1)ex
3
1
−2/3 x
=x
e x(x + 1) + x + (x + 1)
3
1
7
x2 + 3 x + 3
= ex
x2/3
which you might also simplify further to a term with yucky exponents.
Remark 3.21. There is an “anti-product rule”, which is called “integration by parts”; but we’ll spend a whole
lecture on that next month!
3.3.3
The quotient rule
Theorem 3.22. Let f and g be differentiable functions on a common domain A. Then the quotient function
differentiable everywhere that g(x) 6= 0, and its derivative is given by
g(x)f ′ (x) − f (x)g ′ (x)
d f (x)
=
dx g(x)
g(x)2
f
g
is
Note that unlike the product rule, which was symmetric in f and g, this rule is certainly NOT symmetric!
Even the order of the terms in the numerator is vitally important.
Proof. Let’s work on the difference quotient; we use the same clever trick as in the product rule.
1 f (x + h)g(x) − f (x)g(x + h)
1 f (x + h) f (x)
−
=
h g(x + h)
g(x)
h
g(x)g(x + h)
1 f (x + h)g(x) − f (x)g(x) + f (x)g(x) − f (x)g(x + h)
=
h
g(x)g(x + h)
f (x + h) − f (x)
g(x + h) − g(x)
1
g(x)
− f (x)
=
g(x)g(x + h)
h
h
Taking the limit as h → 0, we derive the familiar formula.
Example 3.23.
(x + 1)(2x) − x2 (1)
2x2 + 2x − x2
x2 + 2x
d x2
=
=
=
.
dx x + 1
(x + 1)2
(x + 1)2
(x + 1)2
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3.3.4
Derivatives of the remaining trigonometric functions
Example 3.24. Find the derivative of y = tan(x).
Solution: We recall that tan(x) = sin(x)/ cos(x). Therefore we apply the quotient rule.
d sin(x)
cos(x) cos(x) − sin(x)(− sin(x))
=
dx cos(x)
(cos(x))2
cos2 (x) + sin2 (x)
=
cos2 (x)
1
=
cos2 (x)
= sec2 (x)
Example 3.25. Find the derivative of y = sec(x).
Solution: We recall that sec(x) = 1/ cos(x). Therefore we apply the quotient rule.
1
cos(x) · 0 − 1(− sin(x))
d
=
dx cos(x)
(cos(x))2
1 sin(x)
=
cos(x) cos(x)
= sec(x) tan(x)
We can differentiate csc(x) and cot(x) in a similar way. We deduce four new differentiation formulas:
d
d
(tan(x)) = sec2 (x)
and
(cot(x)) = − csc2 (x)
dx
dx
d
d
(sec(x)) = sec(x) tan(x)
and
(csc(x)) = − csc(x) cot(x)
dx
dx
Remark 3.26. Notice that of the 6 trig functions, the ones whose derivatives have a formula with a minus sign
are the ones that start with a “c”: cos(x), cot(x) and csc(x).
Given these derivatives, we can compute some integrals involving trigonometric functions, but not all.
R π/4
Example 3.27. Find −π/4 sec2 (x) dx.
d
tan(x) = sec2 (x) so
Solution: we sketch the graph and expect a positive answer. We have dx
π/4
Z π/4
2
sec (x) dx = tan(x)
= tan(π/4) − tan(−π/4) = 1 − (−1) = 2.
−π/4
−π/4
Remark 3.28. But BEWARE! What’s wrong with the following?
3π/4
Z 3π/4
2
sec (x) dx = tan(x)
= tan(3π/4) − tan(π/4) = −1 − (1) = −2?
π/4
π/4
Sketch the graph: this area was unbounded, because there was a vertical asymptote in the middle of it; and a
negative answer is patently ridiculous, since if your integrand is the square of something, it’s obviously greater
than or equal to 0. We got a ridiculous answer because we asked a ridiculous question!
A table of the integrals we know now includes:
Z
Z
2
sec (x) dx = tan(x) + c
sec(x) tan(x) dx = sec(x)+c
and
Z
csc2 (x) dx = − cot(x) + c
and
Z
csc(x) cot(x) dx = − csc(x)+c
We memorize the derivatives (and thus anti-derivatives) of all of the basic trigonometric functions.
43