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]