Question

Determine the nth differential coefficient of x^na^x

Answer 1

Answer:

Step-by-step explanation:

$\tt{y=x^n\cdot\,a^x}$

According to Leibniz rule,

For a function y = U.V , then its $n^{th}$ differential coefficient is

$\bf{y_{n}=U_{n}+\,^{n}C_{1}\,U_{n-1}\cdot\,V_{1}+\,^{n}C_{2}\,U_{n-2}\cdot\,V_{2}+\,^{n}C_{3}\,U_{n-3}\cdot\,V_{3}+\dots}$

$\bf{\displaystyle\implies\,y_{n}=\sum^{n}_{k=0}\,^{n}C_{k}\cdot\,U_{n-k}\cdot\,V_{k}}$

$\sf{Put\,\,\,U=a^x\,\,\,\,and\,\,\,\,V=x^n}$

So,

$\bf{\displaystyle\implies\,y_{n}=\sum^{n}_{k=0}\,^{n}C_{k}\cdot\,\dfrac{d}{dx^{n-k}}(a^x)\cdot\,\dfrac{d}{dx^{k}}(x^n)}$

$\bf{\displaystyle\implies\,y_{n}=\sum^{n}_{k=0}\,^{n}C_{k}\cdot\,\dfrac{a^x}{\left(\ln(a)\right)^{n-k}}\cdot\,k!\cdot\,x^{n-k}}$

$\bf{\displaystyle\implies\,y_{n}=\sum^{n}_{k=0}\dfrac{n!}{k!(n-k)!}\cdot\,\dfrac{a^x}{\left(\ln(a)\right)^{n-k}}\cdot\,k!\cdot\,x^{n-k}}$

$\bf{\displaystyle\implies\,y_{n}=\sum^{n}_{k=0}\dfrac{n!}{(n-k)!}\cdot\,\dfrac{a^x}{\left(\ln(a)\right)^{n-k}}\cdot\,x^{n-k}}$