Ask
your own question, for FREE!
Mathematics
10 Online
OpenStudy (christos):
Do you find this differentiation approach sane?
http://screencast.com/t/W1bJgZMz
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
The last derivative you take is wrong.
d/dx (4 + (3x)^1/2) should just be sqrt(3)/(2sqrt(x))
OpenStudy (amistre64):
sane? yes .. alot better than "first principles"
OpenStudy (amistre64):
its the chain rule that i was trying to describe to you last time :)
OpenStudy (christos):
So it's 100% correct?
OpenStudy (amistre64):
well, the sanity of it is 100% ... the correctness along the way may be suspect
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (christos):
inbefore saying that sanity is a derivate of correctness :D
OpenStudy (christos):
derivative*
OpenStudy (amistre64):
\[let:~\\u=4+(3x)^{1/2}\\u'=\frac{3}{2(3x)^{1/2}}\]
\[(4+(3x)^{1/2})^{1/2}~\to~u^{1/2}\]
\[D_x[u^{1/2}]=\frac{1}{2u^{1/2}}*u'\]
\[\frac{1}{2(4+(3x)^{1/2})}*\frac{3}{2(3x)^{1/2}}\]
OpenStudy (christos):
its correct?
OpenStudy (amistre64):
\[\frac{1}{2(4+(3x))^{1/2}}*\frac{3}{2(3x)^{1/2}}\]
\[\frac{3}{4\sqrt{3x} \sqrt{4+(3x)}}\]
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (amistre64):
your "since d/dx" step is off
OpenStudy (amistre64):
can you spot the error by comparing mine with yours?
OpenStudy (christos):
yes I spotted it!
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
clllaaaaaire:
CLOSED
2 weeks ago
0 Replies
0 Medals