Problem Set #3 Solutions

1.a) 42612 23 +−+=′ xxxy

b) 620123636 235 −−−+=′ xxxxy

c)22

2

)42(

12286

+++−=′

x

xxy

d) y’ = ..50( ).10x 2 4 .3x2 12

...12( ).10x 2 5 .3x2 1 x

OR when factored

...32 .3 x2

1 .135x2 .12x 25 ( ).5 x 1

4

OR when expanded

.8100000x8 .1680000x6 .664000x4 .7200000x7 .3168000x5 .209152x3 .105600x2 .15616x 800

e) y’ = ..sec( )θ tan( )θ csc( )θ ..sec( )θ csc( )θ cot( )θ

OR when simplified to cos

Y’ = .2 cos( )θ

21

.cos( )θ2

cos( )θ2

1

f) y’ = 1 + tan2θ OR y’ = sec2θ∴

g) 755 +=′ xey

h) y’ =

which simplifies to

y’ =

..1

3

.6x

( ).4 x 2.12

x2

( ).4 x 22

x2

( ).4 x 2

.2

x

( )x 1

( ).2 x 1

h) In this problem, it is useful to square both sides of the equation before

carrying out the implicit differentiation.

26 42 xyxyy −=

Differentiating, and solving for dy/dx, we get

)43(

)21(5 xxyy

yy

dx

dy

−+−=

j) By the chain rule, we get

y’ =

2. Taking the derivative, we get

cbxaxxf ++= 23)(' 2

We now set this equation to zero and solve using the quadratic formula

a

acbbx

6

1242 2 −±−=

We see that the number of horizontal tangents is dependent on the

number of real solutions.

a) If 0124 2 >− acbThen the quantity under the square root is positive and there aretwo roots (or horizontal tangents)

b) If 0124 2 =− acbThen the quantity under the square root is zero and there is asingle root (-b/3a) and a single horizontal tangent.

..5 x2

4

3

2x

c) If 0124 2 <− acbThen the quantity under the square root is negative, and we have acomplex number which means there are no real roots (or horizontaltangents).

3. If )sin(2)( xxxf +=

Then )cos(21)( xxf +=′

So, the derivative is simply a vertically shifted cosine function with amagnitude of 2. This being the case, we know that it will periodically crossthe x-axis (y=0). Setting the derivative to zero, we get

2

1)cos( −=x and

3

4,

3

2

2

1arccos

ππ=

−

And since this is a periodic function repeating every 2π radians, there arehorizontal tangents at

±

±=

ππ

ππ

n

nx

23

4

23

2

where n is an integer

4. If 1

)(3

−+=

x

xxxf

Then2

23

)1(

132)(

−−−=′

x

xxxf

And so 3(2)=′f

Using the point slope form, the tangent line at the point (2,10) is

y-10=3(x-2) which in standard form is

y=3x+4

5. If 2

6)(

+=

xxf

Then 2)2(

6)(

+−=′

xxf

So, 3

2)1( −=′f

Using the point slope form and simplifying, the tangent line at (1,2) is

3

8

3

2 +−= xy

We know that the slope of the normal to the tangent is the negativereciprocal of the slope of the tangent (3/2 in this case) so using thepoint slope form and simplifying, the normal to the tangent is

2

1

2

3 += xy