<p>Can anyone find the:</p>
<p>y1 = tan(x^x)
y1' = ???</p>
<p>y2 = tan(x)^x
y2' = ???</p>
<p>For y1' i got dy/dx = ((lnx)+1)(arctan(x)(tanx^x)^2 +1) </p>
<p>For y2 i got (tanx)^x (ln(tanx)) + ((arctan^(x-1))x)</p>
<p>Can anyone verify my answers? I doubt they are right ...</p>
<p>y=tan(x^x)
dy/dx=d(x^x)/dx *sec^2(x^x)
The derivative of x^x is found using logarithmic differentiation
z=x^x, lnz=xlnx, dz/dx *1/z=lnx+1, dz/dx=(x^x)(lnx+1)
Plugging this back in, we get dy/dx=(x^x)(1+lnx)sec^2(x^x)</p>
<p>y=tan(x)^x, lny=xln(tan(x)), 1/y(dy/dx)=[x<em>sec^2(x)/tan(x)]+ln(tan(x)) so
dy/dx=(tan(x)^x)[x</em>sec^2(x)/tan(x)+ln(tan(x)). You can have fun cleaning up this last expression :)</p>