Calculus: The Language of Change

Exercise Set 6.4
The Chain Rule

by Michelle Nguyen

Exercise 1 : Drill on Chains

Break  each of the following expressions into a composition of two functions and apply the Chain Rule to find (dy/dx). (The  letters a, b, c denote  constants  or parameters,  and e is the  natural  base for logs and exponentials.)

a

y= (((1 + x))^33)

Solution  of part a

\(\(\(WalkD[\((1 + x)\)\^33, x]; \)
\)\)

b

u = (((a + bx))^33)

WalkD[(((a + b*x))^33),x];

c

y = (((a + b*x))^c)

WalkD[(((a + b*x))^c),x];

d

u =(e^(a x))

WalkD[(E^(a*x)), x];

e

y =(e^((x^2) + 3))

Solution of part e

\(\(\(WalkD[E\^\(x\^2 + 3\), x]; \)

\)\)

f

u = (e^(a*(x^2) + b))

WalkD[ (E^(a*(x^2) + b)), x];

g)

y =(e^Cos)[x]

Solution  of part g

WalkD[(E^Cos)[x],x];

h)

y = (e^Sin)[a*x] + b

WalkD[(E^Sin)[a*x] + b,x];

i)

y=(e^Log)[x]

Solution  of part i

WalkD[(E^Log)[x],x];

j)

y=3((( Sin[x]))^3)

Solution  of part j

WalkD[3(((Sin[x]))^3),x];

k)

y=3 Sin[(x^3)]

Solution  of part k

WalkD[3 Sin[(x^3)],x];

l)

y=3((( Sin[(x^3)]))^3)

Solution  of part l

WalkD[3(((Sin[(x^3)]))^3),x];

m)

y=Log[x(x^(1/5))]

Solution  of part m

WalkD[Log[x (x^(1/5))],x];

n)

y= ((6 Log[x])/5)

Solution  of part n

WalkD[((6 Log[x])/5),x];

o)

y= Log[(x^((6/5)))]

Solution  of part o

WalkD[Log[(x^((6/5)))],x];

p)

y= Log[(x^c)]

Solution  of part p

WalkD[Log[(x^c)],x];

q)

y= c Log[x]

Solution  of part q

WalkD[c Log[x], x];

r)

y= Cos[7 (e^x)]

Solution  of part r

WalkD[ Cos[7 (E^x)],x];

2)

The Product  Sin[θ]Cos[θ]= (1/2)Sin[2θ].  Show this using the addition formula for the sine.  (HINT: Sin[θ+θ]=? )  Use the Chain Rule to show that the derivative of the product is Cos [2θ] by differentiating  the right side of the  equality .  Check your work using the Product Rule on the left side of the equality.

Solution

First we need to prove that	Sin[θ] Cos[θ]=(1/2)\(Sin[2  θ] . 
Then\)  we  use  the product rule and the chain rule on the left and right hand side respectively to show that the derivative of  both sides is \(Cos[2  θ] . \ \ 

                \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
		
	To\) show  Sin[θ]Cos[θ]= (1/2)Sin[2  θ]        we use the identity		 \(Sin\ [θ + θ] = \ 
          \(Sin\ [θ] Cos[θ]\  + Sin[θ] Cos[θ]

            	\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 
            2\ Sin[θ] Cos[θ]\)\)So		Sin[θ + θ] = 
          Sin[2  θ]  =  2 Sin[θ] Cos[θ]Dividing by 2 gives us		Sin[θ] Cos[θ] = (1/2) Sin[2 θ]

\(TextForm
    \`\(Next\ we\ use\ the\ Chain\ Rule\ to\ find\ the\ derivative\ of\ 1\/2\ 
      Sin[2\ θ]
			
\)\)

\(\(WalkD[1\/2\ Sin[2\ θ], \ θ]
\)\)

\(TextForm
    \`Finally\ we\ use\ the\ Product\ Rule\ on\ \ Sin[θ] 
      Cos[θ]\)

\(\(
WalkD[Sin[θ] Cos[θ], θ]\)\)

Does Cos[θ]^2 - Sin[θ]^2 = Cos[2 θ] ? Yes, as we showed in Exercise Set 6.3 number 4.

3)

The  sine function in degrees can be thought  of this way:

                   sIN[D]=Sin[(π/180)D]
  
where Sin[u] denotes the radian measure sine function.  Use  the Chain Rule to show that  the  derivative of the sine in degrees is (π/180)  times the cosine in degrees.

                   y = sIN[D] ⇒ dy/dD = (π/(180 )) cOS[D]

Solution

\(We\ know\ the\ \ Sine\ function\ in\ degrees\ is\ defined\ 
            \(as\  : 
				sIN[D]\)\)=Sin[(π/180)D].
 That  implies  that the  cOS[D]function  in degrees is defined \(as : 
				cOS[D]\)= Cos [(π/180)D].
		

Let             u=  [(π/180)D] ⇒ (du/dD)=(π/180)
				y= sIN[D]=Sin[(π/180)D]  = Sin[u]  ⇒ 
				(dy/du)= Cos[u]   = Cos  [(π/180)D]
                                    We know       (dy/dD)=(dy/du) * (du/dD)So                       (dy/dD)= (π/180)*  Cos [(π/180)D]But                     Cos [(π/180)D] =cOS[D]Hence,         (dy/dD)= (π/180)* cOS [D]

4)

Which parts of Problem 6.2 can you now do using the Chain rule (in addition to the other rules) that  you could not do without it?

Solution

Using the Chain Rule we can now do Problem 6.2 j and k.

j)	 y = (Sqrt[Sin])[x]

\(\(
WalkD[\@Sin[x], x]\)\)

k)		y = Sin[(Sqrt[x])]

WalkD[Sin[(Sqrt[x])],x]