K9-47G Posted October 15, 2005 Share Posted October 15, 2005 1) Find [math] \frac{d}{dx} log(lnx) [/math] I assume that the log has a base of 10, so I got [math] \frac{1}{x(lnxln10)} [/math] 2) Find the slope of the line tangent to the graph [math] cos(xy)=y [/math] at [math] (0,1) [/math] [math] -sin(xy)(y)+(xy')=y' [/math] [math] -ysin(xy)=y'-(xy') [/math] [math] \frac{-ysin(xy)}{1-x}=y' [/math] Then I just keep getting 0 when I substitute (0,1) in... 3) If [math] y=(lnx)^{sinx} x>1, [/math] Find [math] y' [/math] [math] sinxlnx=sinx\frac{1}{x}+(cosx)(lnx) [/math] [math] \frac{sinx}{x} +cosxlnx [/math] [math] 1+cosxlnx [/math] Link to comment Share on other sites More sharing options...

K9-47G Posted October 16, 2005 Author Share Posted October 16, 2005 Can you tell if those answers are right? Link to comment Share on other sites More sharing options...

TD Posted October 16, 2005 Share Posted October 16, 2005 1 is correct, your answer for 2 is correct (slope is 0 at that point, it has a very nice graph btw) and 3 doesn't seem right to me, but it's not really clear what you did. Link to comment Share on other sites More sharing options...

Dave Posted October 16, 2005 Share Posted October 16, 2005 You can make the latex much nicer by using \ln, \log, \sin and \cos btw. For example: [math] \frac{d}{dx} \log(\ln x) [/math] (click to view). Link to comment Share on other sites More sharing options...

cosine Posted October 16, 2005 Share Posted October 16, 2005 <snip>2) Find the slope of the line tangent to the graph [math] cos(xy)=y [/math] at [math] (0' date='1) [/math'] [math] -sin(xy)(y)+(xy')=y' [/math] [math] -ysin(xy)=y'-(xy') [/math] [math] \frac{-ysin(xy)}{1-x}=y' [/math] Then I just keep getting 0 when I substitute (0,1) in... <snip> Sorry, even though the answer is correct, there is a mistake in the work. [math] \cos(xy)=y [/math] [math]-\sin(xy)(y)dx - \sin(xy)(x)dy = dy [/math] [math]-\sin(xy)(y)dx = dy(1 + \sin(xy)(x))[/math] [math]\frac{dy}{dx} = \frac{-\sin(xy)(y)}{1 + \sin(xy)(x)}[/math] That is the derivitive, though for this particular question it just happens to give you the same answer. Hope this helps. Link to comment Share on other sites More sharing options...

K9-47G Posted October 17, 2005 Author Share Posted October 17, 2005 Ok, thanks a lot. Link to comment Share on other sites More sharing options...

K9-47G Posted October 17, 2005 Author Share Posted October 17, 2005 For number 3, I thought I would use the logarithmic power rule (not sure of the real name) and therefore the exponent, sinx, can be written as the first term in problem. Then I used the product rule to find the derivative.. [math] y= (\ln x)^{\sin x} [/math] is the same as [math] \sin x\ln x [/math] Link to comment Share on other sites More sharing options...

BobbyJoeCool Posted October 17, 2005 Share Posted October 17, 2005 you have a problem though... [math](\ln x)^{\sin x} \neq \sin x \cdot \ln x[/math] [math]\ln (x^{\sin x})=\sin x \cdot \ln x[/math] it's not the same thing... the rule is [math]\log _b (x^n) = n \log _b x[/math] [math](\log _b x)^n \neq n \log _b x[/math] Link to comment Share on other sites More sharing options...

BobbyJoeCool Posted October 17, 2005 Share Posted October 17, 2005 You can make the latex much nicer by using \ln' date=' \log, \sin and \cos btw. For example: [math'] \frac{d}{dx} \log(\ln x) [/math] (click to view). What all will that work for? Link to comment Share on other sites More sharing options...

Ducky Havok Posted October 17, 2005 Share Posted October 17, 2005 Might I add, just so you don't mess up with it in the future, that [math]\frac{\sin{x}}{x}[/math] does not always equal 1. [math]\lim_{x\to0}\frac{\sin{x}}{x}=1[/math], but when its not that it's just a regular function and does not simplify. Link to comment Share on other sites More sharing options...

BobbyJoeCool Posted October 19, 2005 Share Posted October 19, 2005 Might I add, just so you don't mess up with it in the future, that [math]\frac{\sin{x}}{x}[/math] does not always equal 1. [math]\lim_{x\to0}\frac{\sin{x}}{x}=1[/math'], but when its not that it's just a regular function and does not simplify. I saw a proof on [math]\lim_{x\to0}\frac{\sin{x}}{x}=1[/math] having something to so with sin(x)<x<tan(x) and then it uses the squeese theorm to say that sin(x)/x as x->0 =1... Link to comment Share on other sites More sharing options...

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