Calculus: The Language of Change

Exercise Set 6.6
Derivative of the Natural Log

by Michelle Nguyen

Exercise 1

For  y [t] as follows, find (dy/dt)[t]

a

y= (((Log[t]))^3)

Solution  of part a

WalkD[(((Log[t]))^3), t];

\(TextForm\`
If\ \ x\ is\ independent\ of\ t, \ then\ \ dx\/dt = 0\)If x is a function of t,  
    there is no effect on this problem  because there is no x in
       the equation .

b

y= Log[Cos[x]]

Solution of part b

\(TextForm\`
If\ x\ is\ independent\ of\ t, \ then\ dx\/dt\  = \ 0\)\(TextForm\`and				dy\/dt\  = \ \(\(dy\/dx\) dx\/dt\  = \ 0\)
\)\(TextForm\`If\ x\ is\ a\ function\ of\ t, \ 
    then\ we\ first\ find \(\( dy\/dx\) : \)
\)

\(\(
\(WalkD[Log[Cos[x]], x]; \)\)\)

\(TextForm
    \`Now			dy\/dt\  = \ 
      \(\(dy\/dx\) dx\/dt\  = \ \ \(-Tan[x]\)\ dx\/dt\)\)

c

y = t Log[t]  -  t

WalkD[ t Log[t]  -  t,t];

\(TextForm\`
If\ \ x\ is\ independent\ of\ t, \ then\ \ dx\/dt = 0\)If x is a function of t,  
    there is no effect on this problem  because there is no x in
       the equation .

d

y= Log[Log[x]]

\(TextForm\`
If\ x\ is\ independent\ of\ t, \ then\ dx\/dt\  = \ 0\)\(TextForm\`and				dy\/dt\  = \ \(\(dy\/dx\) dx\/dt\  = \ 0\)
\)If x is a function of t,  
    then we first find  (dy/dx) :

\(\(\(WalkD[\ Log[Log[x]]\ , x]; \)
\)\)

\(TextForm
    \`Now			dy\/dt\  = \ 
      \(\(dy\/dx\) dx\/dt\  = \ 1\/\(x\ Log[x]\)\ dx\/dt\)\)

e

y =  Log[(t^(((1/t)))) ]

Solution of part e

WalkD[ Log[(t^(((1/t)))) ] , t];

\(TextForm\`
If\ \ x\ is\ independent\ of\ t, \ then\ \ dx\/dt = 0\)If x is a function of t,  
    there is no effect on this problem  because there is no x in
       the equation .

f

y = Log [(t^2)+ 2x]

Solution of part f

This equation contains both t and x.  If x is independent of t, then we treat x as a constant when finding (dy/dt):

WalkD[Log [(t^2)+ 2  x]  ,t];

\(If\ x\ is\ a\ function\ of\ t, \ 
        then\ we\ must\ use\ the\ Chain\ Rule . 
We\ are\ given\ that\)				y= Log [(t^2)+ 2  x]Let u = (t^2)+ 2  x  , then   (du/dt)  =  2 t   +  2  (dx/dt)			and  y  =  Log[u]We know that  (dy/du)   =   (1/u)

So	 (dy/dt)  =  (dy/du) (du/dt) =   (1/u)(  2 t   +  2  (dx/dt))
Substituting back for u gives us
	(dy/dt)  = 	(  (1/( (t^2)+ 2  x  )))( 2 t   +  2  (dx/dt))

Exercise 2

Differentiate y= Log[(x^3)] using u=(x^3) and y= Log[u].  Also use the log identity.  Log [(x^p)]=p Log[x] and differentiate  y = Log[(x^3)]= 3 Log[x] without the Chain rule.  Compare  the two answers.

Solution  :

First Method:  Using the Chain Rule:
 
 Let  u= (x^3)  ⇒ (du/dx) = 3 (x^2)
 
 Let  y= Log[u] ⇒ (dy/du)=(1/u) = (1/(x^3))
 
 We know:
 		(dy/dx)= (dy/du) *  (du/dx)
 
 So from the above equations:
 		(dy/dx)= (1/(x^3)) * 3 (x^2)
                                               
   		(dy/dx) = (3/x)

Second Method:   Using the Log Identity:

		Log [(x^p)]=p Log[x]

This  implies that
		y = Log[(x^3)]= 3 Log[x]

So
		(dy/dx)= 3 * (d/dx)(Log[x])

		(dy/dx)= 3 * (1/x)
       
 		(dy/dx)=(3/x)

Conclusion:  Both methods give us the same answer.