help again with 2 problems

[integragirl]

The first one states: find an equation of the tangent line to the curve y=x^lnx at the point (e,e)
The second is a derivative problem log3(log3(5x^4)) and both 3's are sub-threes at the bottom. PLEASE help!

 

[galactus]

The first one states: find an equation of the tangent line to the curve y=x^lnx at the point (e,e)

Find the derivative of \(\displaystyle \L\\x^{ln(x)}\).

Rewrite \(\displaystyle x^{ln(x)}=e^{ln(x)^{2}}\)

Chain rule:

\(\displaystyle \L\\e^{x}dx, x=ln(x)^{2}\)

\(\displaystyle \L\\x^{2}dx, x=ln(x)\)

\(\displaystyle \L\\(e^{x}dx)(x^{2}dx)(ln(x)dx)\)

\(\displaystyle \L\\2e^{x}ln(x)\frac{1}{x}\)

\(\displaystyle \frac{dy}{dx}=\L\\\frac{2e^{ln(x)^{2}}ln(x)}{x}\)



Use the slope-intercept form, \(\displaystyle \L\\y=\frac{d[x^{ln(x)}]}{dx}x + b\), to solve for b by subbing in your given x and y values.

 

[skeeter]

\(\displaystyle y = x^{lnx}\)

ever used logarithmic differentiation ?

\(\displaystyle lny = ln(x^{lnx})\)

\(\displaystyle lny = (lnx)^2\)

\(\displaystyle \frac{d}{dx}[lny = (lnx)^2]\)

\(\displaystyle \frac{y'}{y} = \frac{2lnx}{x}\)

\(\displaystyle y' = y(\frac{2lnx}{x})\)

\(\displaystyle y' = x^{lnx}(\frac{2lnx}{x})\)

\(\displaystyle y'(e) = e(\frac{2}{e}) = 2\)

tangent line equation at (e,e) ...

\(\displaystyle y - e = 2(x - e)\)