If y = frac{1}{1 + x^{n-m} + x^{p-m}} + frac{ | vedclass

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(A) Given $y = \frac{1}{1 + x^{n-m} + x^{p-m}} + \frac{1}{1 + x^{m-n} + x^{p-n}} + \frac{1}{1 + x^{m-p} + x^{n-p}}$.Multiply the numerator and denomin

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(A) Given $y = \frac{1}{1 + x^{n-m} + x^{p-m}} + \frac{1}{1 + x^{m-n} + x^{p-n}} + \frac{1}{1 + x^{m-p} + x^{n-p}}$.Multiply the numerator and denominator of the first term by $x^m$,the second term by $x^n$,and the third term by $x^p$:$y = \frac{x^m}{x^m + x^n + x^p} + \frac{x^n}{x^n + x^m + x^p} + \frac{x^p}{x^p + x^m + x^n}$.Since the denominators are the same,we can combine the terms:$y = \frac{x^m + x^n + x^p}{x^m + x^n + x^p} = 1$.Since $y = 1$ is a constant,its derivative with respect to $x$ is $\frac{dy}{dx} = 0$.Therefore,the value of $\frac{dy}{dx}$ at any point,including $x = e^{m^{n^p}}$,is $0$.

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 $x = e^{m^{n^p}}$ is equal to:

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