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 $\frac{dy}{dx}=$

• A. $1$
• B. $-1$
• C. $0$
• D. None of these.

Answer 1

The correct option is C

$0$

The explanation for the correct option:

Simplification of expression:

The given equation is $y=\frac{1}{1+x^{n-m}+x^{p-m}}+\frac{1}{1+x^{m-n}+x^{p-n}}+\frac{1}{1+x^{m-p}+x^{n-p}}$.

$$\begin{gathered} \Rightarrow y=\frac{1}{x^0+x^{n-m}+x^{p-m}}+\frac{1}{x^0+x^{m-n}+x^{p-n}}+\frac{1}{x^0+x^{m-p}+x^{n-p}} \\ \Rightarrow y=\frac{1}{x^{m-m}+x^{n-m}+x^{p-m}}+\frac{1}{x^{n-n}+x^{m-n}+x^{p-n}}+\frac{1}{x^{p-p}+x^{m-p}+x^{n-p}} \\ \Rightarrow y=\frac{1}{x^{-m}\left(x^m+x^n+x^p\right)}+\frac{1}{x^{-n}\left(x^n+x^m+x^p\right)}+\frac{1}{x^{-p}\left(x^p+x^m+x^n\right)} \\ \Rightarrow y=\frac{x^m}{\left(x^m+x^n+x^p\right)}+\frac{x^n}{\left(x^n+x^m+x^p\right)}+\frac{x^p}{\left(x^p+x^m+x^n\right)} \\ \Rightarrow y=\frac{x^m+x^n+x^p}{\left(x^m+x^n+x^p\right)} \\ \Rightarrow y=1 \end{gathered}$$

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \therefore \frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow \frac{dy}{dx}=0 \end{gathered}$$

Therefore, $\frac{dy}{dx}=0$.

Hence, (C) is the correct option.