Lecture 7- Continuation and Exam Review PDF

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Lecture 7- Continuation and Exam Review...

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18.01 Single Variable Calculus Fall 2006

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Lecture 7

18.01 Fall 2006

Lecture 7: Continuation and Exam Review Hyperbolic Sine and Cosine Hyperbolic sine (pronounced “sinsh”): sinh(x) =

ex − e−x 2

cosh(x) =

ex + e−x 2

Hyperbolic cosine (pronounced “cosh”):

d � ex − e−x� ex − (−e−x ) d = cosh(x) = sinh(x) = dx dx 2 2 Likewise, d cosh(x) = sinh(x) dx d cos(x).) dx

(Note that this is different from Important identity:

cosh2 (x) − sinh2 (x) = 1

Proof: cosh2 (x) − sinh2 (x)

=

cosh2 (x) − sinh2 (x)

=

�2 � x �2 ex + e−x e − e−x − 2 2 � 1 � 2x � 1 � 2x 1 e + 2ex e−x + e−2x − e − 2 + e−2x = (2 + 2) = 1 4 4 4 �

Why are these functions called “hyperbolic”? Let u = cosh(x) and v = sinh(x), then u 2 − v2 = 1 which is the equation of a hyperbola. Regular trig functions are “circular” functions. If u = cos(x) and v = sin(x), then u 2 + 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�

�

=

u� v + uv� (product rule) u� v − uv� (quotient rule) v2

(uv) � u� � v

d f (u(x)) dx

=

f � (u(x)) · u� (x) (chain rule)

=

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 + 3xy 2 = 8 Instead of solving for y and then taking its derivative, just take example, 3y 2y� + 6xyy� + 3y2

=

2

=

(3y + 6xy)y

�

y�

d of the whole thing. In this dx

0

−3y 2 −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 y� =

1 1 =√ cos y 1 − x2

2

Lecture 7

18.01 Fall 2006

Specific differentiation formulas You will be responsible for knowing formulas for the derivatives and how to deduce these formulas from previous information: x n, sin−1 x, tan−1 x, sin x, cos x, tan x, sec x, e x, ln x . For example, let’s calculate

d sec x: dx

d d 1 −(− sin x) sec x = = = tan x sec x dx dx cos x cos2 x You may be asked to find

d d sin x or cos x, using the following information: dx dx sin(h) h cos(h) − 1 lim h →0 h lim

h →0

=

1

=

0

Remember the definition of the derivative: f (x + ∆x) − f (x) d f (x) = lim dx ∆x→0 ∆x

Tying up a loose end d r x , where r is a real (but not necessarily rational) number? All we have done so far dx is the case of rational numbers, using implicit differentiation. We can do this two ways:

How to find

1st method: base e

x xr d r x dx d r x dx

= = = =

ln x

e � ln x �r e = er ln x d r ln x d r e = e r ln x (r ln x) = e r ln x dx dx x �r� r−1 = rx xr x

2nd method: logarithmic differentiation

(ln f )�

=

f

=

ln f

=

�

(ln f )

=

f � = f (ln f )�

=

f� f xr r ln x r x � r� xr = rxr−1 x 3

Lecture 7

18.01 Fall 2006

Finally, in the first lecture I promised you that you’d learn to differentiate anything— even something as complicated as d x tan−1 x e dx So let’s do it! d uv e dx Substituting, d x tan−1 x e dx

d (uv) = e uv(u� v + uv� ) dx

=

euv

=

ex tan

−1

4

x

�

� tan−1 x + x

1 �� 1 + x2...