Question

If $y=\frac{1}{1+x^{n-m}+x^{p-m}}+\frac{1}{1+x^{m-n}+x^{p-n}}+\frac{1}{1+x^{m-p}+x^{n-p}},then\frac{dy}{dx}$ is equal to

Answer 1

$\frac{1}{1+x^{n-m}+x^{p-m}}+\frac{1}{1+x^{m-n}+x^{p-n}}+\frac{1}{1+x^{m-p}+x^{n-p}}=\frac{1}{1+\frac{x^n}{x^m}+\frac{x^p}{x^m}}+\frac{1}{1+\frac{x^m}{x^n}+\frac{x^p}{x^n}}+\frac{1}{1+\frac{x^m}{x^p}+\frac{x^n}{x^p}}=\frac{x^m}{x^m+x^n+x^p}+\frac{x^n}{x^n+x^m+x^p}+\frac{x^p}{x^p+x^m+x^n}=\frac{(x^m+x^n+x^p)}{(x^m+x^n+n^x)}=1\therefore \frac{dy}{dx}=0$