Question

If \[y = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .....\], then find the value of \[\dfrac{{dy}}{{dx}}\]

Answer 1

Hint: A function of infinite number of terms is given. We have to find its differentiation. So, differentiate all the terms of the function which are given and find the value of the given factorials. After that simplify the obtained series. The series after differentiation will be same as the given series.

Formula Used:
The factorial of \[n\] is \[n! = n\left( {n - 1} \right)\left( {n - 2} \right).....2 \cdot 1\]
The differentiation of the sum of functions is equal to the sum of the differentiation of the functions i.e.
\[\dfrac{d}{{dx}}\left\{ {{f_1}\left( x \right) + {f_2}\left( x \right) + ..... + {f_n}\left( x \right)} \right\} = \dfrac{d}{{dx}}\left\{ {{f_1}\left( x \right)} \right\} + \dfrac{d}{{dx}}\left\{ {{f_2}\left( x \right)} \right\} + ..... + \dfrac{d}{{dx}}\left\{ {{f_n}\left( x \right)} \right\}\]
The differentiation of \[{x^n}\] is \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]

Complete step-by-step answer:
The given equation is \[y = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + .....\]
The factorial of \[n\] is \[n! = n\left( {n - 1} \right)\left( {n - 2} \right).....2 \cdot 1\]
Putting \[n = 1\], we get \[1! = 1\]
Putting \[n = 2\], we get \[2! = 2 \cdot 1 = 2\]
Putting \[n = 3\], we get \[3! = 3 \cdot 2 \cdot 1 = 6\]
So, the given equation becomes \[y = 1 + x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{6} + .....\]
Differentiating both sides of the above equation, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left\{ {1 + x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{6} + .....} \right\}\]
Differentiating the terms of the right-hand side one by one, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( 1 \right) + \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{2}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{x^3}}}{6}} \right) + ..... - - - - - \left( i \right)\]
Differentiation of any constant term is zero.
So, \[\dfrac{d}{{dx}}\left( 1 \right) = 0\]
and the differentiation of \[{x^n}\] is \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Putting \[n = 1\], we get \[\dfrac{d}{{dx}}\left( x \right) = 1{x^{1 - 1}} = {x^0} = 1\]
Putting \[n = 2\], we get \[\dfrac{d}{{dx}}\left( {{x^2}} \right) = 2{x^{2 - 1}} = 2{x^1} = 2x\]
Putting \[n = 3\], we get \[\dfrac{d}{{dx}}\left( {{x^3}} \right) = 3{x^{3 - 1}} = 3{x^2}\]
Substituting these values in equation \[\left( i \right)\], we get
\[\dfrac{dy}{dx}=0+1+\dfrac{2x}{2}+\dfrac{3x^{2}}{6}+\cdots\]
Simplify this equation.
\[\dfrac{dy}{dx}=1+x+\dfrac{x^{2}}{2}+\cdots\]
This is the same expression as the expression for \[y\].
Hence the value of \[\dfrac{{dy}}{{dx}}\] is equal to \[y\].

Note:- Here the right hand side expression of the given equation is the sum of an infinite number of terms. So, we differentiate each and every term individually using the rule which states that “The differentiation of the sum of functions is equal to the sum of the differentiation of the functions”. After finding the differentiation we compare the obtained expression to the given expression.