- #1
Logarythmic
- 281
- 0
Problem:
[tex]u(x) = u_1(x)u_2(x)\cdot\cdot\cdot u_n(x)[/tex]
Show that
[tex]u'(x) = u(x)\left[\frac{u'_1(x)}{u_1(x)} + \frac{u'_2(x)}{u_2(x)} + ... + \frac{u'_n(x)}{u_n(x)}\right][/tex]
Hint: Take the logarithm of u(x) first.
I solved this by using the chain rule. Regarding the hint, am I missing something?
Next problem states that, by using the formula for u'(x), differentiate
[tex]u(r) = \frac{1}{(r+1)(r+2)\cdot\cdot\cdot(r+n)}[/tex]
When using the formula, my answer to this is
[tex]u'(r) = -\frac{1}{(r+1)(r+2)\cdot\cdot\cdot(r+n)} \left[\frac{1}{r+1} + \frac{1}{r+2} + ... + \frac{1}{r+n}\right][/tex]
but the correct answer should be
[tex]u'(r) = (r+1)(r+2)\cdot\cdot\cdot(r+n)\left[\frac{1}{r+1} + \frac{1}{r+2} + ... + \frac{1}{r+n}\right][/tex]
This seems like an elementary problem but I'm not getting the correct result so I must be missing something..?
[tex]u(x) = u_1(x)u_2(x)\cdot\cdot\cdot u_n(x)[/tex]
Show that
[tex]u'(x) = u(x)\left[\frac{u'_1(x)}{u_1(x)} + \frac{u'_2(x)}{u_2(x)} + ... + \frac{u'_n(x)}{u_n(x)}\right][/tex]
Hint: Take the logarithm of u(x) first.
I solved this by using the chain rule. Regarding the hint, am I missing something?
Next problem states that, by using the formula for u'(x), differentiate
[tex]u(r) = \frac{1}{(r+1)(r+2)\cdot\cdot\cdot(r+n)}[/tex]
When using the formula, my answer to this is
[tex]u'(r) = -\frac{1}{(r+1)(r+2)\cdot\cdot\cdot(r+n)} \left[\frac{1}{r+1} + \frac{1}{r+2} + ... + \frac{1}{r+n}\right][/tex]
but the correct answer should be
[tex]u'(r) = (r+1)(r+2)\cdot\cdot\cdot(r+n)\left[\frac{1}{r+1} + \frac{1}{r+2} + ... + \frac{1}{r+n}\right][/tex]
This seems like an elementary problem but I'm not getting the correct result so I must be missing something..?