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Logarithmic Differentation
- Thread starter nikchic5
- Start date
Try rewriting \(\displaystyle \L\\x^{ln(x)}\) as \(\displaystyle e^{(ln(x))^{2}}\)
Use the chain rule.
\(\displaystyle \L\\\frac{d}{dx}[e^{(ln(x))^{2}}]\)
Take the derivative of \(\displaystyle \L\\(ln(x))^{2}=\frac{2ln(x))}{x}\)
Since \(\displaystyle \L\\x^{ln(x)}=e^{(ln(x))^{2}}\), we have:
\(\displaystyle \L\\y'=\frac{2ln(x)x^{ln(x)}}{x}\)
Use the chain rule.
\(\displaystyle \L\\\frac{d}{dx}[e^{(ln(x))^{2}}]\)
Take the derivative of \(\displaystyle \L\\(ln(x))^{2}=\frac{2ln(x))}{x}\)
Since \(\displaystyle \L\\x^{ln(x)}=e^{(ln(x))^{2}}\), we have:
\(\displaystyle \L\\y'=\frac{2ln(x)x^{ln(x)}}{x}\)
nikchic5 said:If anyone could help me with this one it would be great!
Use logarithmic differentation to find the derivative.
y= (x)^(ln x)
If anyone could help me by showing me step-by-step it would be greatly appreicated!! Thank you soo much!
1. Take the Ln of both sides and simplify:
\(\displaystyle Ln(y) = Ln(x^{Lnx})\)
\(\displaystyle Ln(y) = Ln(x)Ln(x)\)
\(\displaystyle Ln(y) = Ln^2(x)\)
2. Differentiate implicitly and simplify:
\(\displaystyle \L\\ \frac{1}{y}\frac{dy}{dx} = \frac{2Ln(x)}{x}\)
\(\displaystyle \L\\\frac{dy}{dx} = y[\frac{2Ln(x)}{x}]\)
but, \(\displaystyle y=x^{Lnx}\)
\(\displaystyle \L\\ \frac{dy}{dx} = [x^{Lnx}]\frac{2Ln(x)}{x}\)