We have $$(1-\ln x)^2\ge0$$ and, moreover, $(1-\ln x)^2=0\iff \ln x=1$. Can you find the solution of $\ln x =1$?
#3
markosheehan
133
0
how do you get the derivative of it
#4
Evgeny.Makarov
Gold Member
MHB
2,434
4
As for any composition of functions. The outer function if $f(x)=x^2$, so first you differentiate $x^2$, and the result is $2x$. Now our whole function is not $f(x)$, but $f(1-\ln x)$. Therefore you replace $x$ in $2x$ with $1-\ln x$ to get $2(1-\ln x)$. To finish, you need to multiply this by the derivative of $1-\ln x$.
I recommend reviewing the chain rule for computing the derivative of the composition of functions in your textbook.
I'm reviewing Meirovitch's "Methods of Analytical Dynamics," and I don't understand the commutation of the derivative from r to dr:
$$
\mathbf{F} \cdot d\mathbf{r} = m \ddot{\mathbf{r}} \cdot d\mathbf{r} = m\mathbf{\dot{r}} \cdot d\mathbf{\dot{r}}
$$