MATH 1A SECTION: OCTOBER 11, 2013
Consider a function f (x) with derivative f 0 (x). Recall that the value f 0 (a) of the derivative
at a point a is the slope of the tangent line to f (x) at x = a.
1. Where does f (x) = ex − 2x have a horizontal tangent line?
2. Can y = 2ex + 3x + 5x3 have a tangent line with slope 2?
3. Find real numbers a and b such that y = ax2 + bx has y = 3x − 2 as a tangent line at
(1, 1).
4. Find y 0 if x3 + y 3 = 1.
√
√
5. Find y 0 if 2 x + y = 3.
6. Find
dy
dx
7. Find
dy
dx
if ey cos x = 1 + sin(xy).
√
if x + y = 1 + x2 y 2 .
8. Find y 0 if arctan(x2 y) = x + xy 2 .
Solutions and Commentary
Recently we have been emphasizing the importance of turning mathematics into English
– writing mathematics in complete sentences with capital letters and periods, making sure
to communicate ideas. We started off today with a few problems about translating English
into mathematics – given a problem stated in words, how do we properly turn that into a
math problem that we can solve?
1. Let f (x) = ex − 2x. This function has a horizontal tangent line exactly when the slope
of the tangent line is zero, so when f 0 (x) = 0. Note that f 0 (x) = ex − 2, so we need to
solve ex − 2 = 0. This gives ex = 2, and hence x = ln 2 is the only value where f (x) has
a horizontal tangent line.
2. y = 2ex + 3x + 5x3 has derivative y 0 = 2ex + 3 + 15x2 . This function has a tangent line
of slope 2 exactly when y 0 = 2. We are therefore looking for solutions to the equation
2ex + 3 + 15x2 = 2. We can rewrite this as
2ex + 15x2 = −1.
Now, observe that ex > 0 for all x (that is, (0, ∞) is the range of ex ). Also, x2 ≥ 0 for
all x. Therefore, 2ex + 15x2 is the sum of two non-negative quantities, which definitely
cannot equal a negative number such as −1. Therefore, y 0 = 2 has no solutions, and hence
y = 2ex + 3x + 5x3 cannot have a tangent line of slope 2.
3. Let f (x) = ax2 + bx. Then f 0 (x) = 2ax + b. Note that for f (x) to have a tangent line at
(1, 1), it must pass through (1, 1), so f (1) = 1. Also, the value of the derivative f 0 (1) is
the slope of the tangent line to f (x) at 1; this tangent line is y = 3x − 2 with slope 3, so
therefore f 0 (1) = 3.
Therefore, we have a system of two equation in two unknowns:
f (1) = 1
f 0 (1) = 3.
Plugging x = 1 into our expressions for f (x) and f 0 (x), we can rewrite this system of
equations as
a+b=1
2a + b = 3.
We must now solve this system. Subtract the first equation from the second equation
to get a = 2; substituting then yields b = −1. Therefore, we have shown that y = 2x2 − 1
has y = 3x − 2 as a tangent line at (1, 1).
Warning: It is very tempting to solve this problem by solving for f 0 (x) = 3x − 2, and
this was a common mistake today. This is unfortunately not a correct method to solve
this problem. Keep in mind that the value of the derivative only gives the slope of the
tangent line, and not the entire tangent line.
This problem is tricky – it’s a quadratic whose derivative is linear, but this linear
function for the derivative is not the linear function that is the tangent line. If instead we
were working with a function whose derivative was not linear, this problem would likely
be much less confusing.
2
Now, we switch to thinking about problems involving implicit differentiation.
4. If x3 + y 3 = 1, we can differentiate to see that
3x2 + 3y 2 y 0 = 0.
Solving for y 0 yields 3y 2 y 0 = −3x2 , so therefore
y0 = −
x2
3x2
=
−
.
3y 2
y2
d
Note that we think about differentiating dx
(y 3 ) by using the chain rule. We would
d
3
2
usually differentiate dx (x ) = 3x ; now, we have a function y nested inside, so we get
d
d
(y 3 ) = 3y 2 dx
(y) = 3y 2 y 0 . Observe that here, we are thinking of y as a function of x,
dx
and not as a constant.
√
√
5. We have 2 x + y = 3. Differentiate to get
1
1
√ + √ y 0 = 0.
x 2 y
We can arrange this to become
1
1
√ y0 = − √ ,
2 y
x
so therefore
√
2 y
y =−√ .
x
0
6. We have ey cos x = 1 + sin(xy). We can differentiate, carefully applying the product rule
and chain rule as needed:
so therefore
rearranging yields
so therefore
d y
d
d
(e ) cos x + ey (cos x) = cos(xy) (xy)
dx
dx
dx
dy
dy
y
y
e
cos x − e sin x = cos(xy) y + x
dx
dx
dy
dy
ey cos x − x cos(xy)
= ey sin x + y cos(xy)
dx
dx
dy
ey sin x + y cos(xy)
= y
.
dx
e cos x − x cos(xy)
3
√
7. We have x + y = 1 + x2 y 2 . We can differentiate, applying the product rule and chain
rule. This gives:
d
d
d
1
√
(x + y) = x2 (y 2 ) + (x2 )y 2
2 x + y dx
dx
dx
1
dy
dy
√
which is
1+
= x2 2y
+ 2xy 2
2 x+y
dx
dx
dy
dy
1
1
√
·
− 2x2 y
= 2xy 2 − √
rearranging gives
2 x + y dx
dx
2 x+y
1
2
2xy − 2√x+y
dy
so hence
.
= 1
√
dx
− 2x2 y
2 x+y
8. We have arctan(x2 y) = x+xy 2 . Recall that
gives
which is (applying the product rule)
which rearranges to become
d
(arctan x)
dx
=
1
.
1+x2
Therefore, differentiating
1
d
d
· (x2 y) = 1 + (xy 2 )
2
2
1 + (x y) dx
dx
1
2xy + x2 y 0 = 1 + y 2 + 2xyy 0
4
2
1+x y
2xy
x2
y 0 − 2xyy 0 = 1 + y 2 −
.
4
2
1+x y
1 + x4 y 2
Therefore, we can conclude that
0
y =
1 + y2 −
x2
1+x4 y 2
2xy
1+x4 y 2
− 2xy
,
which can be simplified, but that simplification will not be shown here.
4