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Math 3A Homework #4 solutions
4.2:4 Differentiate f ( x ) = − 3 x^4 + 6 x^2 − 2.
Answer:
d
dx
[− 3 x
4
2 − 2 ] = (− 3 )( 4 ) x
3
= − 12 x^3 + 12 x
4.2:26 Differentiate f ( x ) = a^2 x^4 − 2 ax^2.
Answer:
d
dx
[ a^2 x^4 − 2 ax^2 ] = ( a^2 )( 4 ) x^3 − ( 2 a )( 2 ) x
= 4 a
2 x
3 − 4 ax
4.2:44 Find the line tangent to y = − 2 x^3 − 3 x + 1 at x = 1.
Answer: The slope of the tangent line is
m =
dy
dx
= − 6 x
2 − 3
When x = 1, m = − 6 ( 1 )^2 − 3 = −9 and f ( 1 ) = − 2 ( 1 )^3 − 3 ( 1 ) + 1 = −4. Using the point-slope formula, we
see
( y − (− 4 )) = − 9 · ( x − 1 )
which yields y = − 9 x + 5 for the tangent line. We can verify this graphically:
0
5
10
15
-1 -0.5 0 0.5 1 1.5 2 2.5 3
-2x3 -3x+ -9*x+
4.2:50 Find the line normal to y = 1 − 3 x^2 at x = −2.
Answer: The slope of the tangent line is
m =
dy
dx
= − 6 x
When x = −2, m = − 6 (− 2 ) = 12 and f (− 2 ) = 1 − 3 (− 2 )^2 = −11. The normal line has a slope that is the
negative reciprocal of the tangent line’s slope, or − 1 /12. Using the point-slope formula, we see
( y − (− 11 )) = −
· ( x − (− 2 ))
which yields y = − 121 x − 676 for the normal line.
4.3:2 Differentiate f ( x ) = ( 2 x − 1 )( 2 − x^2 ).
Answer: Using the product rule
d
dx
f ( x ) = ( 2 x − 1 )
d
dx
( 2 − x
2 ) + ( 2 − x
2 )
d
dx
( 2 x − 1 )
= ( 2 x − 1 )(− 2 x ) + ( 2 − x^2 )( 2 )
= − 6 x
2
4.3:48 Differentiate R ( x ) = k ( a − x )( b − x ).
Answer: Using the product rule
d
dx
f ( x ) = k ( a − x )
d
dx
( b − x ) + k ( b − x )
d
dx
( a − x )
= k ( a − x )(− 1 ) + k ( b − x )(− 1 )
= k ( 2 x − a − b )
4.3:50 Differentiate f ( x ) = 3 x
(^2) − 2 x + 1 2 x + 1.
Answer: By the quotient rule
f
′ ( x ) =
( 2 x + 1 )( 6 x − 2 ) − ( 3 x^2 − 2 x + 1 )( 2 )
( 2 x + 1 )^2
6 x^2 − 2 x
( 2 x + 1 )^2
4.3:78 Differentiate f ( x ) =
ax^2 k^2 + x^2.
Answer: By the quotient rule
f ′( x ) =
( k^2 + x^2 )( 2 ax ) − ( ax^2 )( 2 x )
( k^2 + x^2 )^2
2 k^2 ax
( k^2 + x^2 )^2
4.4:8 Differentiate h ( x ) =
5 x + 3 x^2.
Answer: Let g ( x ) = 5 x + 3 x^2 and f ( u ) =
u. Then g ′( x ) = 5 + 6 x and f ′( u ) = 1 2
√ u
. So then
h
′ ( x ) = f
′ ( g ( x )) · g
′ ( x )
5 + 6 x
2
5 x + 3 x^2
4.4:18 Differentiate h ( s ) =
2 s s + 1
At ( 1 ,
3 2
3 ), we get the slope of the tangent line to be m = −
9 4
1 3 2
√ 3
√ 3
- Then the slope of the normal
line is √^2 3
Use the point slope formula to get the equations for the tangent
( y −
( x − 1 )
y =
x + 2
and the normal
( y −
( x − 1 )
y =
x +
We can confirm these by plotting them:
0
1
2
3
4
5
-2 -1 0 1 2
