Math 127H: Lecture 8
Trig functions and Implicit Differentiation
Trig Functions
We give the two facts
•
d
(sin(x))
dx
= cos(x)
•
d
(cos(x)
dx
= − sin(x)
Later we will prove these. For now we concentrate on how to use these facts.
Example 1.
d
(7x2 − 1) sin(x) ⇒ (use the product rule) ⇒
dx
d
d
( (7x2 − 1)) sin(x) + (7x2 − 1)( sin(x)) = 7x sin(x) + (7x2 − 1) cos(x). (1)
dx
dx
Example 2. We take the derivative of sin(3x2 − 1). We use the chain rule. This function
is the composition
x −→ (3x2 − 1) = u −→ sin(u).
Hence
d
d
d
(sin(3x2 − 1) =
(3x2 − 1) (sin(u)) = 6x cos(x).
dx
dx
du
Example 3. We use the power rule to take the derivative
d
(cos(x))1/2 = (1/2) cos(x)−1/2 (− sin(x)).
dx
Implicit Differentiation
We say that a function y = y(x) is defined explicitly if it is given in the form
y = f (x).
We say it is given implicitly if it given as a relation
f (x, y) = g(x, y).
We give a couple of examples:
x2 − 3y 3 = 1
3x2 y − 4xy = x + y.
If y is given implicitly it is an honest function of x and can be treated with the usual
dy
rules of differentiation. This allows us to find dx
in terms of x and y.
1
Example 4. From the relation x2 − 3y 2 = 1 by differentiating with respect to x we obtain
dy
=0
dx
dy
2x = 6y
dx
dy
= x/3y.
dx
2x − 6y
Inverse Functions and Derivatives
We say that f, g are inverse functions provided f ◦ g = I. Here I denotes the identity
function, that is, the function I(x) = x. The derivative of the identity function is 1. We
have
x 7→ g(x) = u 7→ f (u) = x.
The last equality is the assumption that f, g are inverses to each other. Apply the chain
rule. We have
dx
dg df
=
= 1.
dx du
dx
dg
df
If we know two of dx
or du
we can figure out the third from the above equality.
Example 5. We write the function g(x) = ex as g(x) = exp(x). The two functions exp
and ln are inverses to one another, that is,
ln ◦ exp = I,
x → exp(x) = u → ln(u) = x.
Apply the chain rule to the above equation. One gets
d
d
d
(exp(x)) (ln(u)) = exp(x) (ln(u)) = 1.
dx
du
du
Solving for the derivative of ln gives
1/exp(x) =
1/u =
d
(ln(u)),
du
d
(ln(u)).
du
Here we have made the substitution exp(x) = u because we want to have the derivative of
ln(u) in terms of u.
Example 6. We find the derivative of ln(u) in a slightly different way. We also have that
exp ◦ ln = I.
This says that
x → ln(x) = u → exp(u) = x.
2
Apply the chain rule to this. One gets
d
d
(exp(u)) (ln(x) = 1
du
dx
d
exp(u) (ln(x) = 1
dx
d
(ln(x)) = 1/exp(u) = 1/x.
dx
We have used the fact that exp(u) = exp(ln(x)) = x.
√
Example 7. Let f (x) = x2 = u, g(u) = u so
g ◦ f = I,
or
√
x → f (x) = x2 = u → u = x.
√
We use this to find the derivative of u. Applying the chain rule we find
d
d
(f ) (g) = 1
dx
du
d 2 d √
(x ) ( u) = 1
dx
dx
d √
2x ( u) = 1
du
d √
( u) = 1/2x
du
√
d √
( u) = 1/(2 u).
du
3