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

Answer: 0

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 ${\left(a^b\right)}^c=a^{bc}$ , 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,
$\frac{\left(l+m\right)}{\left(m-n\right)\left(n-l\right)}+\frac{\left(m+n\right)}{\left(n-l\right)\left(l-m\right)}+\frac{\left(n+l\right)}{\left(l-m\right)\left(m-n\right)}$
$=\frac{\left(l+m\right)(l-m)+(m+n)(m-n)+(n+l)(n-l)}{\left(m-n\right)\left(n-l\right)(l-m)}$
$=\frac{\left(l^2-m^2\right)+\left(m^2-n^2\right)+\left(n^2-l^2\right)}{\left(m-n\right)\left(n-l\right)\left(l-m\right)}$
$=0$
Hence, $P=x^0$ .
$P=1$
$\frac{dP}{dx}=\frac{d}{dx}\left(1\right)$
$P^{{}^{\prime }}\left(x\right)=0$
Hence, the differential coefficient of $P$ is $0$ .