Jump to content

Derivatives: Some Questions...


K9-47G

Recommended Posts

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

<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

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

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

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

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.