Question

Use logarithmic differentiation to calculate the derivative.
F(x)=x^sqrtx
I need step by step! I have worked out solution but I don't understand.

Answer 1

\[y=x^\sqrt{x}\]
Take natural log of both sides!
\[\ln(y)=\ln(x^{\sqrt{x}})\]
Using law of log from algebra we can rewrite right hand side so that we have:
\[\ln(y)=\sqrt{x} \ln(x)\]
Now to find y' we need to differentiate both sides.
\[[\ln(y)]'=\frac{y'}{y}\]
\[[\sqrt{x} \ln(x)]'=[\sqrt{x}]' \ln(x)+\sqrt{x} [\ln(x)]'=\frac{1}{2 \sqrt{x}} \ln(x)+\sqrt{x} \frac{1}{x}\]

Answer 2

So we have
\[\frac{y'}{y}=\frac{1}{2 \sqrt{x}} \ln(x)+\sqrt{x} \frac{1}{x}\]
Now multiply y on both sides!
\[y'=y (\frac{1}{2 \sqrt{x}} \ln(x)+\sqrt{x} \frac{1}{x})\]
Then replace y with x^sqrt{x}
\[y'=x^{\sqrt{x}} (\frac{1}{2 \sqrt{x}} \ln(x)+\sqrt{x} \frac{1}{x})\]

Answer 3

there answer is \[2+lnx \div 2\sqrt{x} \times x ^{\sqrt{x}}\]

Answer 4

I have up to what you have but I dont understand how to get there answer

Answer 5

The two answers are exactly the same; just write out my's term in parentheses with a common denominator of 2sqrt(x)

Answer 6

thanks

Answer 7

And if I can give you some friendly notational advice, try to avoid using the "divided by sign" now you're in calculus. It is problematic and avoided in preference for writing expressions as fractions.
i.e., instead of
\[a \div b \]
write a/b or
\[{a \over b}\]