The differential coefficient of {left( {{x^{f | vedclass

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(B) Let the given expression be $y = {\left( {{x^{\frac{{\ell + m}}{{m - n}}}}} \right)^{\frac{1}{{n - \ell }}}} \cdot {\left( {{x^{\frac{{m + n}}{{n

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(B) Let the given expression be $y = {\left( {{x^{\frac{{\ell + m}}{{m - n}}}}} ight)^{\frac{1}{{n - \ell }}}} \cdot {\left( {{x^{\frac{{m + n}}{{n - \ell }}}}} ight)^{\frac{1}{{\ell - m}}}} \cdot {\left( {{x^{\frac{{n + \ell }}{{\ell - m}}}}} ight)^{\frac{1}{{m - n}}}}$.Using the property of exponents $(x^a)^b = x^{ab}$,we have:$y = x^{\frac{\ell + m}{(m - n)(n - \ell)}} \cdot x^{\frac{m + n}{(n - \ell)(\ell - m)}} \cdot x^{\frac{n + \ell}{(\ell - m)(m - n)}}$.Since the bases are the same,we add the exponents:$y = x^{\left[ \frac{\ell + m}{(m - n)(n - \ell)} + \frac{m + n}{(n - \ell)(\ell - m)} + \frac{n + \ell}{(\ell - m)(m - n)} ight]}$.To add these fractions,we find a common denominator,which is $(m - n)(n - \ell)(\ell - m)$.Multiplying the numerator of each term by the missing factor:Exponent $= \frac{(\ell + m)(\ell - m) + (m + n)(m - n) + (n + \ell)(n - \ell)}{(m - n)(n - \ell)(\ell - m)}$.Using the identity $(a+b)(a-b) = a^2 - b^2$:Exponent $= \frac{(\ell^2 - m^2) + (m^2 - n^2) + (n^2 - \ell^2)}{(m - n)(n - \ell)(\ell - m)} = \frac{0}{(m - n)(n - \ell)(\ell - m)} = 0$.Thus,$y = x^0 = 1$ for $x 
eq 0$.The derivative of a constant $1$ with respect to $x$ is $\frac{d}{dx}(1) = 0$.

The differential coefficient of ${\left( {{x^{\frac{{\ell + m}}{{m - n}}}}} \right)^{\frac{1}{{n - \ell }}}} \cdot {\left( {{x^{\frac{{m + n}}{{n - \ell }}}}} \right)^{\frac{1}{{\ell - m}}}} \cdot {\left( {{x^{\frac{{n + \ell }}{{\ell - m}}}}} \right)^{\frac{1}{{m - n}}}}$ with respect to $x$ is:

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