Math 105: Review for Exam II - Solutions
1. Find dy/dx
dy/dx
dy/dx for each of the following.
(a) y=x2+ 2x+e2+e2x+ ln 2 + ln(2x) + arctan 2
y=x2+ 2x+e2+e2x+ ln 2 + ln(2x) + arctan 2
y=x2+ 2x+e2+e2x+ ln 2 + ln(2x) + arctan 2
dy
dx = 2x+ (ln 2)2x+ 2e2x+1
2x·2 Note that e2, ln 2, and arctan 2 are constants.
(b) y=√x·arctan(5x)
y=√x·arctan(5x)
y=√x·arctan(5x)
dy
dx =1
2x−1/2arctan(5x) + √x·1
1 + (5x)2·5 = arctan(5x)
2x1/2+5√x
1 + 25x2
(c) y= ln(tan(2cos(x2)))
y= ln(tan(2cos(x2)))
y= ln(tan(2cos(x2)))
dy
dx =1
tan(2cos(x2))·sec2(2cos(x2))·ln 2(2cos(x2))·(−sin(x2)) ·2x
(d) y=x+eπ
cos 4 + sin5(6x)
y=x+eπ
cos 4 + sin5(6x)
y=x+eπ
cos 4 + sin5(6x)Note that eπand cos 4 are constants.
dy
dx =(1)(cos 4 + sin5(6x)) −(x+eπ)(5 sin4(6x)·cos(6x)·6)
(cos 4 + sin5(6x))2Recall that sin5(6x) = (sin(6x))5.
2. Consider the curve defined by x3+y3=9
2xy
x3+y3=9
2xy
x3+y3=9
2xy (known as the Folium of Descartes).
(a) Find dy/dx
dy/dx
dy/dx.Use implicit differentiation.
3x2+ 3y2dy
dx =9
2y+9
2xdy
dx
3y2dy
dx −9
2xdy
dx =9
2y−3x2
dy
dx 3y2−9
2x=9
2y−3x2
dy
dx =
9
2y−3x2
3y2−9
2x
(b) Verify that the point (1,2) is on the curve above.
We must check to see if the values x= 1 and y= 2 satisfy the equation above.
x3+y3?
=9
2xy
13+ 23?
=9
2·1·2
9?
= 9
Thus, the point (1,2) is on the curve.
(c) Find the equation of the tangent line at the point (1,2).
We want y=mx +b.
m=
9
2·2−3·12
3·22−9
2·1=4
5, so y=4
5x+b.
Now plug in x= 1 and y= 2 to find b.
2 = 4
5·1 + b⇒6
5=b
Therefore, we have y=4
5x+6
5.
