Question

Let f be a differentiable function that is positive everywhere and such that

f′(x)=−f(x)ln(f(x))for allx∈ℝ.

f′(x)=−f(x)ln⁡(f(x))for allx∈R.

(a) Use logarithmic differentiation for the function g(x)=(f(x))^e^x to show that g′(x)=0 for all x∈ℝ

(b) Then, using the fact that any function whose derivative vanishes over some interval must be constant over that interval, deduce that f(x)=e^(e^-x)* ln C for some positive constant C

Answer 1

$\mathrm{(}a\mathrm{)}\ \ g\left(x\right)={\left(f\left(x\right)\right)}^{e^x} \\ \space Taking\ \ \mathrm{"ln"}\ \ of\ \ both\ \ side, \\ \\ \\ \mathrm{ln}\left(g\left(x\right)\right)=e^x\ \mathrm{ln}\left(f\left(x\right)\right) \\ \\ Differentiate\ Both\ \ Side\ \ of\ \ the\ \ equation, \\ \\ \frac{g'\left(x\right)}{g\left(x\right)}\ \ =\ e^x\ \mathrm{ln}\left(f\left(x\right)\right)+e^x\ \frac{f'\left(x\right)}{f\left(x\right)} \\ \\ g'\left(x\right)=g\left(x\right)\left(e^x\ \mathrm{ln}\left(f\left(x\right)\right)+e^x\ \frac{f'\left(x\right)}{f\left(x\right)}\right) \\ \\ But\ \ given\ \ that\ \ \ f'\left(x\right)=-f\left(x\right){}{}{}\ \ \mathrm{ln}\left(f\left(x\right)\right) \\ \\ Therefore,\ \ g'\left(x\right)=g\left(x\right)\left(e^x\ \mathrm{ln}\left(f\left(x\right)\right)-e^x\ \frac{f\left(x\right){}{}{}\ \ \mathrm{ln}\left(f\left(x\right)\right)}{f\left(x\right)}\right) \\ \\ g'\left(x\right)=g\left(x\right)\left(e^x\ \mathrm{ln}\left(f\left(x\right)\right)-e^x\ \mathrm{ln}\left(f\left(x\right)\right)\right)\ \ =\ \ 0 \\ \\ \left(b\right)f'\left(x\right)=-f\left(x\right){}{}{}\ \ \mathrm{ln}\left(f\left(x\right)\right) \\ \\ \frac{df\left(x\right)}{ \begin{array}{l} f\left(x\right){}{}{}\ \ \mathrm{ln}\left(f\left(x\right)\right) \\ \end{array} }=-dx \\ \\ \int{\frac{d\left(f\left(x\right)\right)}{f\left(x\right){}{}{}\ \ \mathrm{ln}\left(f\left(x\right)\right)}=-x+\mathrm{ln}C} \\ \\ \mathrm{ln}\left(\mathrm{ln}\left(f\left(x\right)\right)\right)=-x+\mathrm{ln}C \\ \\ \mathrm{ln}\left(f\left(x\right)\right)=e^{-x+C} \\ \\ f\left(x\right)=e^{\mathrm{(}e^{-x}\mathrm{)ln}C}$