If \[y = \dfrac{1}{{1 + {x^{n - m}} + {x^{p - m}}}} + \dfrac{1}{{1 + {x^{m - n}} + {x^{p - n}}}} + \dfrac{1}{{1 + {x^{m - p}} + {x^{n - p}}}}\] then \[\dfrac{{dy}}{{dx}}\] is equals to
A) 1
B) -1
C) 0
D) None of these
Answer
528.9k+ views
Hint: Here we can observe that the second and third term of every denominator is with a power that ends with the same variable. This is the hint we will use. Along with that, we will use the very common rule of indices and then take the LCM of the denominator. Then we will come to know that the denominators of all the terms are the same so we will add them together and will take the derivative.
Complete step by step solution:
Given that,
\[y = \dfrac{1}{{1 + {x^{n - m}} + {x^{p - m}}}} + \dfrac{1}{{1 + {x^{m - n}} + {x^{p - n}}}} + \dfrac{1}{{1 + {x^{m - p}} + {x^{n - p}}}}\]
We know that,
\[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\]
Applying this in the denominator we can write,
\[y = \dfrac{1}{{1 + \dfrac{{{x^n}}}{{{x^m}}} + \dfrac{{{x^p}}}{{{x^m}}}}} + \dfrac{1}{{1 + \dfrac{{{x^m}}}{{{x^n}}} + \dfrac{{{x^p}}}{{{x^n}}}}} + \dfrac{1}{{1 + \dfrac{{{x^m}}}{{{x^p}}} + \dfrac{{{x^n}}}{{{x^p}}}}}\]
Now taking LCM of the denominator terms in the denominators we get,
\[y = \dfrac{1}{{\dfrac{{{x^m} + {x^n} + {x^p}}}{{{x^m}}}}} + \dfrac{1}{{\dfrac{{{x^n} + {x^m} + {x^p}}}{{{x^n}}}}} + \dfrac{1}{{\dfrac{{{x^p} + {x^m} + {x^n}}}{{{x^p}}}}}\]
Shift the denominator of the denominator to the numerator we get,
\[y = \dfrac{{{x^m}}}{{{x^m} + {x^n} + {x^p}}} + \dfrac{{{x^n}}}{{{x^n} + {x^m} + {x^p}}} + \dfrac{{{x^p}}}{{{x^p} + {x^m} + {x^n}}}\]
Now since the denominators are same we can directly add the numerators,
\[y = \dfrac{{{x^m} + {x^n} + {x^p}}}{{{x^m} + {x^n} + {x^p}}}\]
Numerator and denominator of the fraction are same so cancelling them we get,
\[y = 1\]
Now the value of the given function is a constant. And we know that the derivative of a constant is always zero. Thus we can say that,
\[\dfrac{{dy}}{{dx}} = 0\]
Thus option (C) is the correct option.
Note:
Note that this problem seems to be very lengthy but actually it uses only one rule of indices and powers. So don’t try any other way to solve this. Also note that the powers are already arranged in such a way that the final y function turns out to be 1.
Note that the rules are different for \[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\] and \[{a^{m + n}} = {a^m}{a^n}\].
Complete step by step solution:
Given that,
\[y = \dfrac{1}{{1 + {x^{n - m}} + {x^{p - m}}}} + \dfrac{1}{{1 + {x^{m - n}} + {x^{p - n}}}} + \dfrac{1}{{1 + {x^{m - p}} + {x^{n - p}}}}\]
We know that,
\[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\]
Applying this in the denominator we can write,
\[y = \dfrac{1}{{1 + \dfrac{{{x^n}}}{{{x^m}}} + \dfrac{{{x^p}}}{{{x^m}}}}} + \dfrac{1}{{1 + \dfrac{{{x^m}}}{{{x^n}}} + \dfrac{{{x^p}}}{{{x^n}}}}} + \dfrac{1}{{1 + \dfrac{{{x^m}}}{{{x^p}}} + \dfrac{{{x^n}}}{{{x^p}}}}}\]
Now taking LCM of the denominator terms in the denominators we get,
\[y = \dfrac{1}{{\dfrac{{{x^m} + {x^n} + {x^p}}}{{{x^m}}}}} + \dfrac{1}{{\dfrac{{{x^n} + {x^m} + {x^p}}}{{{x^n}}}}} + \dfrac{1}{{\dfrac{{{x^p} + {x^m} + {x^n}}}{{{x^p}}}}}\]
Shift the denominator of the denominator to the numerator we get,
\[y = \dfrac{{{x^m}}}{{{x^m} + {x^n} + {x^p}}} + \dfrac{{{x^n}}}{{{x^n} + {x^m} + {x^p}}} + \dfrac{{{x^p}}}{{{x^p} + {x^m} + {x^n}}}\]
Now since the denominators are same we can directly add the numerators,
\[y = \dfrac{{{x^m} + {x^n} + {x^p}}}{{{x^m} + {x^n} + {x^p}}}\]
Numerator and denominator of the fraction are same so cancelling them we get,
\[y = 1\]
Now the value of the given function is a constant. And we know that the derivative of a constant is always zero. Thus we can say that,
\[\dfrac{{dy}}{{dx}} = 0\]
Thus option (C) is the correct option.
Note:
Note that this problem seems to be very lengthy but actually it uses only one rule of indices and powers. So don’t try any other way to solve this. Also note that the powers are already arranged in such a way that the final y function turns out to be 1.
Note that the rules are different for \[{a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}\] and \[{a^{m + n}} = {a^m}{a^n}\].
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