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 $\frac{dy}{dx}$ at $e^{m^{n^p}}$ is equal to

• A. $e^{mnp}$
• B. $e^{mn/p}$
• C. $e^{np/m}$
• D. $0$

Answer 1

The correct option is D $0$

$y=\frac{1}{1+x^{n-m}+x^{p-m}}+\frac{1}{1+x^{m-n}+x^{p-n}}+\frac{1}{1+x^{m-p}+x^{n-p}}\Rightarrow y=\frac{1}{1+\frac{x^n}{x^m}+\frac{x^p}{x^m}}+\frac{1}{1+\frac{x^m}{x^n}+\frac{x^p}{x^n}}+\frac{1}{1+\frac{x^m}{x^p}+\frac{x^n}{x^p}}$

$\Rightarrow y=\frac{x^m}{x^m+x^n+x^p}+\frac{x^n}{x^m+x^n+x^p}+\frac{x^p}{x^m+x^n+x^p}$

$=\frac{1}{(x^m+x^n+x^p)}(x^m+x^n+x^p)$

$\Rightarrow y=1\Rightarrow \frac{dy}{dx}=0$ at all $x$

$\therefore \frac{dy}{dx}$ at $e^{m^{n^p}}=0$

D answer