Logarithmic differentiation (how to solve log differentiation with two different term)

Question

f(m)=m log m+(n-m)log (n-m)

f'(m)=log m +1+(-1)log (n-m)+(-1). How to get this line?

f'(n/2)=0

f(n/2) is the minimum point. How to know n/2 is the minimum point?

Answer 1

You have, as you wrote, $$f(m)=m \log (m)+(n-m) \log (n-m)$$ $$f'(m)=\log (m)-\log (n-m)$$ $$f''(m)=\frac{1}{n-m}+\frac{1}{m}$$ So, the serivative cancels when $$\log (m)=\log (n-m)\implies m=n-m\implies 2m=n\implies m=\frac n 2$$ Now, using the second derivative test $$f''\left(\frac n 2\right)=\frac{4}{n}>0$$ So, the point is the minimum and $$f\left(\frac n 2\right)=n \log \left(\frac{n}{2}\right)$$