Continuation and Exam Review

Hyperbolic Sine and Cosine
Hyperbolic sine (pronounced “sinsh”):
sinh(x)= ex ? e?x
2 Hyperbolic cosine (pronounced “cosh”): ex + e?x
cosh(x)=
2
xx
d sinh(x)= de? e?x = e? (?e?x) = cosh(x)
dx dx 22 Likewise,
d
cosh(x) = sinh(x)
dx d
(Note that this is different from cos(x).)
dx
Important identity:
cosh2(x) ? sinh2(x)=1
Proof:
??2 ? x ?2 cosh2(x) ? sinh2(x)= ex +2e?x ? e?2e?x 1 ?? 1 ?? 1cosh2(x) ? sinh2(x)= 4 e2x +2exe?x + e?2x ? 4 e2x ? 2+ e?2x = 4(2+2)=1
Why are these functions called “hyperbolic”? Let u = cosh(x) and v = sinh(x), then u2 ? v2 =1 which is the equation of a hyperbola. Regular trig functions are “circular” functions. If u = cos(x) and v = sin(x), then u2 + v2 =1 which is the equation of a circle.
1
? ?
? ?
Lecture 7 18.01 Fall 2006
Exam 1 Review General Differentiation Formulas
(u + v)? = u? + v? (cu)? = cu? (uv)? = u?v + uv? (product rule)
u ? = u?v ? uv? (quotient rule)
vv2 d
f(u(x)) = f?(u(x)) u?(x) (chain rule)
dx ·
You can remember the quotient rule by rewriting
u
? =(uv?1)?
v
and applying the product rule and chain rule.
Implicit differentiation
Let’s say you want to find y? from an equation like y3 +3xy2 =8 d
Instead of solving for y and then taking its derivative, just take of the whole thing. In this
dx
example, 3y2y? +6xyy? +3y2 =0 (3y2 +6xy)y? = ?3y2 y? = ?3y2
3y2 +6xy
Note that this formula for y? involves both x and y. Implicit differentiation can be very useful for taking the derivatives of inverse functions.
For instance,
y = sin?1 x sin y = x
?
Implicit differentiation yields (cos y)y? =1
and
11
y? ==
cos y ?1 ? x2
2