Question

Differential coefficient of $\left\{x^{\frac{l + m}{m - n}}\right\}^{\frac{1}{n - l}}\cdot \left\{x^{\frac{m + n}{n - l}}\right\}^{\frac{1}{l - m}}\cdot \left\{x^{\frac{n + l}{l - m}}\right\}^{\frac{1}{m - n}}$ with respect to $x$ is:

Answer 1

Let P = $\left(\left\{x^{\left(\frac{l + m}{m - n}\right)}\right\}\right)^{\frac{1}{n - l}}\left(\left\{x^{\left(\frac{m + n}{n - l}\right)}\right\}\right)^{\frac{1}{l - m}}\left(\left\{x^{\left(\frac{n + l}{l - m}\right)}\right\}\right)^{\frac{1}{m - n}}$
As $\left(a^{b}\right)^{c}=a^{b c}$ , we get,
$P=\left\{x^{\left(\frac{\left(l + m\right)}{\left(m - n\right) \left(n - l\right)}\right)}\right\}\left\{x^{\left(\frac{\left(m + n\right)}{\left(n - l\right) \left(l - m\right)}\right)}\right\}\left\{x^{\left(\frac{\left(n + l\right)}{\left(l - m\right) \left(m - n\right)}\right)}\right\}$
As $\left(a^{b}\right)\left(a^{c}\right)=a^{b + c}$ , we get,
$P=x^{\left\{^{\frac{\left(l + m\right)}{\left(m - n\right) \left(n - l\right)} + \frac{\left(m + n\right)}{\left(n - l\right) \left(l - m\right)} + \frac{\left(n + l\right)}{\left(l - m\right) \left(m - n\right)}}\right\}}$
Solving the exponent of $x$ we get,
$\frac{\left(l + m\right)}{\left(m - n\right) \left(n - l\right)}+\frac{\left(m + n\right)}{\left(n - l\right) \left(l - m\right)}+\frac{\left(n + l\right)}{\left(l - m\right) \left(m - n\right)}$
$=\frac{\left(l + m\right) \left(\right. l - m \left.\right) + \left(\right. m + n \left.\right) \left(\right. m - n \left.\right) + \left(\right. n + l \left.\right) \left(\right. n - l \left.\right)}{\left(m - n\right) \left(n - l\right) \left(\right. l - m \left.\right)}$
$=\frac{\left(l^{2} - m^{2}\right) + \left(m^{2} - n^{2}\right) + \left(n^{2} - l^{2}\right)}{\left(m - n\right) \left(n - l\right) \left(l - m\right)}$
$=0$
Hence, $P=x^{0}$ .
$P=1$
$\frac{d P}{d x}=\frac{d}{d x}\left(1\right)$
$P^{'}\left(x\right)=0$
Hence, the differential coefficient of $P$ is $0$ .