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Question: The \({n^{th}}\) derivative of \(x{e^x}\) vanishes when \(\left( a \right)\) x = 0 \(\left( b \r...

The nth{n^{th}} derivative of xexx{e^x} vanishes when
(a)\left( a \right) x = 0
(b)\left( b \right) x = – 1
(c)\left( c \right) x = - n
(d)\left( d \right) x = n

Explanation

Solution

In this particular question use the concept that the differentiation of ddxmn=mddxn+nddxm\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m and use the concept that ddxxn=nxn−1,ddxex=ex\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}{e^x} = {e^x} so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given equation
xexx{e^x}
Let, f(x)=xexf\left( x \right) = x{e^x}
Now differentiate the above equation w.r.t x and using the property that ddxmn=mddxn+nddxm\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m so we have,
⇒ddxf(x)=xddxex+exddxx\Rightarrow \dfrac{d}{{dx}}f\left( x \right) = x\dfrac{d}{{dx}}{e^x} + {e^x}\dfrac{d}{{dx}}x
Now as we know that ddxxn=nxn−1,ddxex=ex\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}{e^x} = {e^x} so we have,
⇒f′(x)=xex+ex(1)\Rightarrow f'\left( x \right) = x{e^x} + {e^x}\left( 1 \right)
⇒f′(x)=ex+xex\Rightarrow f'\left( x \right) = {e^x} + x{e^x}
Now again differentiate it w.r.t x we have,
⇒ddxf′(x)=ddx(ex+xex)\Rightarrow \dfrac{d}{{dx}}f'\left( x \right) = \dfrac{d}{{dx}}\left( {{e^x} + x{e^x}} \right)
⇒ddxf′(x)=ddxex+ddxxex\Rightarrow \dfrac{d}{{dx}}f'\left( x \right) = \dfrac{d}{{dx}}{e^x} + \dfrac{d}{{dx}}x{e^x}
⇒f′′(x)=ex+ex+xex\Rightarrow f''\left( x \right) = {e^x} + {e^x} + x{e^x}
⇒f′′(x)=2ex+xex\Rightarrow f''\left( x \right) = 2{e^x} + x{e^x}
Now again differentiate w.r.t x we have,
⇒ddxf′′(x)=ddx(2ex+xex)\Rightarrow \dfrac{d}{{dx}}f''\left( x \right) = \dfrac{d}{{dx}}\left( {2{e^x} + x{e^x}} \right)
⇒f′′′(x)=2ex+ex+xex\Rightarrow f'''\left( x \right) = 2{e^x} + {e^x} + x{e^x}
⇒f′′′(x)=3ex+xex\Rightarrow f'''\left( x \right) = 3{e^x} + x{e^x}
Similarly
.
.
.
⇒fn(x)=nex+xex\Rightarrow {f^n}\left( x \right) = n{e^x} + x{e^x}............. (1)
Now according to the question we have to find out the condition when nth{n^{th}} derivative vanishes.
⇒fn(x)=0\Rightarrow {f^n}\left( x \right) = 0
So from equation (1) we have,
⇒fn(x)=nex+xex=0\Rightarrow {f^n}\left( x \right) = n{e^x} + x{e^x} = 0
⇒nex+xex=0\Rightarrow n{e^x} + x{e^x} = 0
⇒xex=−nex\Rightarrow x{e^x} = - n{e^x}
⇒x=−n\Rightarrow x = - n
So this is the required condition.
Hence option (c) is the correct answer.

Note : Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation property which is stated above then using these properties differentiate the equation n times as above then equate its nth{n^{th}} derivative is zero and simplify as above, we will get the required condition.