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Derivatives: Some Questions...


K9-47G
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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]

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<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.

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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]

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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]

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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...

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