Question

if \(u = {e^{{a_1}{x_1} + }}^{{a_2}{x_2} + .... + {a_n}{x_n}}\) where\({a^2}_1 + a_2^2 + \ldots a_n^2 = 1\), show that

\(\frac{{{\gamma ^2}u}}{{\gamma x_1^2}} + \frac{{{\gamma ^2}u}}{{\gamma x_2^2}} + ...... + \frac{{{\gamma ^2}u}}{{\gamma x_n^2}} = u\)

Short answer

We should know about partial differentiation to prove the problem.

Step-by-step solution

Given data

Given:\(u = {e^{{a_1}{x_1} + }}^{{a_2}{x_2} + .... + {a_n}{x_n}}\)

Differentiatingpartially with respect to x,

\(\frac{{\gamma u}}{{\gamma {x_1}}} = a{e^{{a_1}{x_1} + }}^{{a_2}{x_2} + .... + {a_n}{x_n}}\)

\(\frac{{{\gamma ^2}u}}{{\gamma x_{_1}^2}} = a_1^2{e^{{a_1}{x_1} + }}^{{a_2}{x_2} + .... + {a_n}{x_n}}\)

Deficient with respect to x

\(\frac{{{\gamma ^2}u}}{{\gamma x_2^2}} = a_x^2{e^{{a_1}{x_1} + }}^{{a_2}{x_2} + .... + {a_n}{x_n}}\)

Then, \(\frac{{{\gamma ^2}u}}{{\gamma x_{_1}^2}} + \frac{{{\gamma ^2}u}}{{\gamma x_2^2}} + ......\frac{{{\gamma ^2}u}}{{\gamma x_n^2}}\)

\( = \left( {{a^2}_1 + a_2^2 + \ldots a_n^2} \right){e^{{a_1}{x_1} + }}^{{a_2}{x_2} + .... + {a_n}{x_n}}\)

Substitution:

Since, \({a^2}_1 + a_2^2 + \ldots a_n^2 = 1\) (given)

\(\frac{{{\gamma ^2}u}}{{\gamma x_{_1}^2}} + \frac{{{\gamma ^2}u}}{{\gamma x_2^2}} + ......\frac{{{\gamma ^2}u}}{{\gamma x_n^2}} = {e^{{a_1}{x_1} + }}^{{a_2}{x_2} + .... + {a_n}{x_n}}\)

Therefore, \(\frac{{{\gamma ^2}u}}{{\gamma x_{_1}^2}} + \frac{{{\gamma ^2}u}}{{\gamma x_2^2}} + ......\frac{{{\gamma ^2}u}}{{\gamma x_n^2}} = u\)