<p>(lnx - 1)/(x^2 * x^1/x)</p>
<p>how did you get that? I got something waaay different
i used the derrivative of U^N= NU^(N-1) * DU/DX
which when I substituted became: 1/x * 1/x ^ (1/x - 1) * -1/x^2
am I doing this totally wrongly?? I don't understand..!</p>
<p>Logarithmic differentiation.</p>
<p>y = x^(-1/x)
ln y = -1/x ln x
1/y. dy/dx = 1/x^2. lnx - 1/x^2
dy/dx = (lnx - 1)/(x^2 * x^1/x)</p>
<p>hey thanks so much. I've been slacking on Calc (AB) guess I have lots of studying to do.</p>
<p>sorry one last question:
using the tangent line what can you tell about 1 / sqrt of 10?</p>
<p>ughh... 1/SQRT(10) is a line itself... I do not understand your question.</p>
<p>I didn't understand it either.<br>
It's okay, thanks for all your help!</p>
<p>Set z= 1/x, then y = z^z</p>
<p>You need dy/dx =(dy/dz).(dz/dx)
Since y = z^z, ln(y) = z ln(z); differentiate both sides wrt z
(1/y) dy/dz = z (1/z) + ln(z) = 1 + ln(z)
or dy/dz = y (1 + ln(z))
Finally, dy/dx = dy/dz . dz/dx = y(1 + ln(1/x)) . (-1)(x^-2)
= - y (1 - ln(x)) / (x^2)</p>
<p>You can plug in y = (1/x)^(1/x) on the RHS, if you want it completely in terms of x .</p>
<p>primeminister - that is another way to do it, but i think in this particular case it would be much quicker using logarithmic differentiation</p>
<p>optimizerdad - i have a question for you, if you don't mind me asking. What is your profession?</p>
<hr>
<p>vrumchev | rumchev.com/forums</p>
<p>primeminister, I have just written some simple notes regarding the technique I used to solve your problem... you can check them out here Notes</a> on Logarithmic Differentiation</p>
<hr>
<p>vrumchev | [url=<a href="http://www.rumchev.com/forums%5Drumchev.com/forums%5B/url">http://www.rumchev.com/forums]rumchev.com/forums[/url</a>]</p>
<p>Hey.. needed help with another problem: </p>
<p>Water runs into a conical tank at the rate of 2 cubic feet per minute. The tank stands point down and has a height of 10 feet and a maximum radius of 5 feet. How fast is the water level rising when the water is 6feet deep?</p>
<p>A) </p>
<p>DV/dt=2</p>
<p>B=Pi r^2
h=2r
volume of cone = 1/3 B.H
V=1/3 (pi.r^2)(2r)
v= 2/3 ( pi. r^3) </p>
<p>now I need to find dv/dt when H=6, r = 3</p>
<p>However, something doesnt seem right. Can anyone please verify? </p>
<p>Thanks much..</p>
<p>P.S: Vmruchev, thanks for the notes... they're great.. :)</p>
<p>Here are a few more questions i'd appreciate some help with:</p>
<p>1) x+sin(y)= Pi/2
1+cosy dy/dx = 0
dy/dx = arccosy -1</p>
<p>2) tan (x+y)=x
x+y = arctan x
1+dy/dx = 1/(1+x^2) . 2x
dy/dx = 2x/(1+x^2) -1</p>
<p>3) cos(x)+cos(y)=2y
-sinx -siny dy/dx = 2
dy/dx = arcsiny (-2-sinx) </p>
<p>4) xy=e^(xy)
xdy/dx+y=e^(xy)
dy/dx = (e^(xy) - y )/( x)</p>
<p>5) taken from my first set of questions:</p>
<p>e^cosy= x^3 arctany
e^(Cosy) . ( -siny) = 3x^2 arctany + 1/1+y^2.dy/dx . x^3
therefore, dy/dx = (e^cosy. (-siny) - 3x^2arctan y . ( 1+y^2) ) / (x^3) </p>
<p>thanks much</p>
<ol>
<li>correct</li>
<li>I would just use sec (same thing-yours is right I think):
tan(x+y)=x
sec(x+y)^2*(1+dy/dx)=1
dy/dx=1/(sec(x+y)^2)-1</li>
<li>I think you missed a dy/dx
cos(x)+cos(y)=2y
-sin(x)-sin(y)dy/dx=2dy/dx
-sin(x)=2dy/dx+sin(y)dy/dx
dy/dx=(-2+sin(y))/sin(x)</li>
<li><p>xy=e^(xy)
x(dy/dx)+y=e^(xy)*(x(dy/dx)+y)
I am a little confused on how to simplify this because the dy/dx is canceling itself out every time....well you forget to mutliply by the derivative of the exponent in e^(xy)</p></li>
<li><p>You missed another dy/dx
e^cosy=x^3arctany
e^(cosy)<em>(-siny)</em>(dy/dx)=(x^3)<em>(1/(y^2+1))</em>(dy/dx)+(arctany)(3x^2)
-(sinye^cosy)<em>(dy/dx)=((x^3)/(y^2+1))</em>(dy/dx)+arctany)(3x^2)
-(3x^2)arctany=((x^3)/(y^2+1))<em>(dy/dx)+sinye^cosy)
dy/dx=-(3x^2</em>arctany)/((x^3/(y^2+1))+sinye^cosy)</p></li>
</ol>
<p>Hope this helps. Just remember the basic rules and you'll be fine.</p>