Math 251
Jonny Comes
Spring 2006
Assignment #8 Partial Solutions
Additional Exercises:
1. Use logarithmic differentiation to show that
(fghk)0=f0ghk +fg0hk +f gh0k+f ghk0.
Solution: Set y=fghk. Then we have
ln(y) = ln(fghk)⇒ln(y) = ln(f) + ln(g) + ln(h) + ln(k).
Now differentiating both sides with respect to xyields
y0
y=f0
f+g0
g+h0
h+k0
k
⇒y0=yf0
f+g0
g+h0
h+k0
k
⇒y0=f ghk f0
f+g0
g+h0
h+k0
k
⇒y0=f ghk f0
f+f ghk g0
g+f ghk h0
h+f ghk k0
k
⇒y0=f0ghk +f g0hk +f gh0k+fghk0
which is exactly what we were trying to show. . .that was fun.
2. Use the result from the previous problem to calculate the derivative of
f(x) = ln(x) sin(x) arctan(x)√x.
Solution: Using the last problem we have
f0(x) = d
dx ln(x)sin(x) arctan(x)√x
+ ln(x)d
dx sin(x)arctan(x)√x
+ ln(x) sin(x)d
dx arctan(x)√x
+ ln(x) sin(x) arctan(x)d
dx √x
=1
xsin(x) arctan(x)√x
+ ln(x) cos(x) arctan(x)√x
+ ln(x) sin(x)1
1+x2√x
+ ln(x) sin(x) arctan(x)−1
√x
=sin(x) arctan(x)
√x+ ln(x) cos(x) arctan(x)√x+ln(x) sin(x)√x
1+x2−ln(x) sin(x) arctan(x)
√x
