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1) Find $\frac{d}{dx} log(lnx)$

I assume that the log has a base of 10, so I got

$\frac{1}{x(lnxln10)}$

2) Find the slope of the line tangent to the graph $cos(xy)=y$ at $(0,1)$

$-sin(xy)(y)+(xy')=y'$

$-ysin(xy)=y'-(xy')$

$\frac{-ysin(xy)}{1-x}=y'$

Then I just keep getting 0 when I substitute (0,1) in...

3) If $y=(lnx)^{sinx} x>1,$ Find $y'$

$sinxlnx=sinx\frac{1}{x}+(cosx)(lnx)$

$\frac{sinx}{x} +cosxlnx$

$1+cosxlnx$

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Can you tell if those answers are right?

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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.

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You can make the latex much nicer by using \ln, \log, \sin and \cos btw. For example:

$\frac{d}{dx} \log(\ln x)$

(click to view).

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<snip>

2) Find the slope of the line tangent to the graph $cos(xy)=y$ at $(0' date='1) [/math'] [math] -sin(xy)(y)+(xy')=y'$

$-ysin(xy)=y'-(xy')$

$\frac{-ysin(xy)}{1-x}=y'$

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.

$\cos(xy)=y$

$-\sin(xy)(y)dx - \sin(xy)(x)dy = dy$

$-\sin(xy)(y)dx = dy(1 + \sin(xy)(x))$

$\frac{dy}{dx} = \frac{-\sin(xy)(y)}{1 + \sin(xy)(x)}$

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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Ok, thanks a lot.

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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..

$y= (\ln x)^{\sin x}$ is the same as $\sin x\ln x$

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you have a problem though...

$(\ln x)^{\sin x} \neq \sin x \cdot \ln x$

$\ln (x^{\sin x})=\sin x \cdot \ln x$

it's not the same thing...

the rule is

$\log _b (x^n) = n \log _b x$

$(\log _b x)^n \neq n \log _b x$

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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?

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Might I add, just so you don't mess up with it in the future, that $\frac{\sin{x}}{x}$ does not always equal 1. $\lim_{x\to0}\frac{\sin{x}}{x}=1$, but when its not that it's just a regular function and does not simplify.

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Might I add, just so you don't mess up with it in the future, that $\frac{\sin{x}}{x}$ does not always equal 1. $\lim_{x\to0}\frac{\sin{x}}{x}=1[/math'], but when its not that it's just a regular function and does not simplify. I saw a proof on [math]\lim_{x\to0}\frac{\sin{x}}{x}=1$ 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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