Find the equation of the tangent to the curve

[Guest]

Find the equation of the tangent to the curve defined by y= x^(x^2) at the point where x=2.

Am I doing this right? : y=x^(x^2)
lny= ln x^x^2
lny= x^2lnx
dy/dx (1/y)= 2xlnx+ x^2(1/x)
dy/dx = 2xlnx+ x^2(1/x) x^x^2

then how do I figure out the equation again?

thanks for the help

 

[pka]

You are generally correct. But please, please use grouping symbols!

\(\displaystyle \L
\begin{array}{rcl}
y & = & x^{x^2 } \\
y' & = & \left( {x^{x^2 } } \right)\left[ {2x\ln (x) + x} \right] \\
y'(2) & = & \left( {2^4 } \right)\left[ {4\ln (2) + 2} \right] \\
\end{array}\)

 

[Guest]

how do you find the equation? because the answer is 32[(1+2(ln2)]x-y-[16(3+8(ln2)]=0

 

[pka]

Well by simple algebra:\(\displaystyle \L
\left( {2^2 } \right)\left[ {4\ln (2) + 2} \right] = \left( 4 \right)\left[ {4\ln (2) + 2} \right] = \left( 8 \right)\left[ {2\ln (2) + 1} \right].\)

 

[Guest]

why is there a 32 in the answer?where did that come from?