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Question: If a polynomial in x is given by \(f\left( x \right)\) is a polynomial in x, then the second derivat...

If a polynomial in x is given by f(x)f\left( x \right) is a polynomial in x, then the second derivative of f(ex)f\left( {{e}^{x}} \right) at x=1x=1 is
(A). ef(e)+f(e)ef''\left( e \right)+f'\left( e \right)
(B). (f(e)+f(e))e2\left( f''\left( e \right)+f'\left( e \right) \right){{e}^{2}}
(C). e2f(e){{e}^{2}}f''\left( e \right)
(D). (f(e)e+f(e))e\left( f''\left( e \right)e+f'\left( e \right) \right)e

Explanation

Solution

Hint: Here, we should know the three important formula of derivative which is ddx(ex)=ex\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}} , ddx[f(g(x))]=g(x)f(g(x))\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=g'\left( x \right)\cdot f'\left( g\left( x \right) \right) ; ddx[g(x)h(x)]=g(x)h(x)+h(x)g(x)\dfrac{d}{dx}\left[ g\left( x \right)\cdot h\left( x \right) \right]=g\left( x \right)\cdot h'\left( x \right)+h\left( x \right)\cdot g'\left( x \right) . We have to find second order derivative of ddx[ddxf(ex)]\dfrac{d}{dx}\left[ \dfrac{d}{dx}f\left( {{e}^{x}} \right) \right] . After getting the final answer we will replace the value of x as 1 wherever x is present in the answer.

Complete step-by-step solution -
Now, here we have to find second order derivative of ddx[ddxf(ex)]\dfrac{d}{dx}\left[ \dfrac{d}{dx}f\left( {{e}^{x}} \right) \right] . So, there are basically 3 formulas which will be used here.
ddx(ex)=ex\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}} ……………………………….(1)
ddx[f(g(x))]=g(x)f(g(x))\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=g'\left( x \right)\cdot f'\left( g\left( x \right) \right) ………………………..(2)
ddx[g(x)h(x)]=g(x)h(x)+h(x)g(x)\dfrac{d}{dx}\left[ g\left( x \right)\cdot h\left( x \right) \right]=g\left( x \right)\cdot h'\left( x \right)+h\left( x \right)\cdot g'\left( x \right) ……………………………..(3)
So, first we will find a first order derivative i.e. ddxf(ex)\dfrac{d}{dx}f\left( {{e}^{x}} \right) . So, we will use here equation(1) and (2) where g(x)=exg\left( x \right)={{e}^{x}} . So, applying the formula we will get
ddxf(ex)=exf(ex)\dfrac{d}{dx}f\left( {{e}^{x}} \right)={{e}^{x}}\cdot f'\left( {{e}^{x}} \right) ………………………….(4)
Now, keeping this value of equation (4) into our main equation to find second derivative i.e.
ddx[exf(ex)]\dfrac{d}{dx}\left[ {{e}^{x}}\cdot f'\left( {{e}^{x}} \right) \right]
So, here using equation (3) as we have 3 terms i.e. g(x)=ex,h(x)=f(ex)g\left( x \right)={{e}^{x}},h\left( x \right)=f'\left( {{e}^{x}} \right)
So, applying the formula we get
ddx[ex.f(ex)]=exf(ex)+ddx[f(ex)]ex\Rightarrow \dfrac{d}{dx}\left[ {{e}^{x}}.f'\left( {{e}^{x}} \right) \right]={{e}^{x}}\cdot f'\left( {{e}^{x}} \right)+\dfrac{d}{dx}\left[ f'\left( {{e}^{x}} \right) \right]\cdot {{e}^{x}} ……………………………(5)
As we can see in the above equation after plus sign there is the term which we have to derive using the formula given in equation (2). So, derivation of that particular term will be
ddx[f(ex)]=exf(ex)\dfrac{d}{dx}\left[ f'\left( {{e}^{x}} \right) \right]={{e}^{x}}\cdot f''\left( {{e}^{x}} \right)
So, substituting this value back into equation (5), we get as
ddx[ex.f(ex)]=exf(ex)+exf(ex)ex\Rightarrow \dfrac{d}{dx}\left[ {{e}^{x}}.f'\left( {{e}^{x}} \right) \right]={{e}^{x}}\cdot f'\left( {{e}^{x}} \right)+{{e}^{x}}\cdot f''\left( {{e}^{x}} \right)\cdot {{e}^{x}}
ddx[ex.f(ex)]=exf(ex)+e2xf(ex)\Rightarrow \dfrac{d}{dx}\left[ {{e}^{x}}.f'\left( {{e}^{x}} \right) \right]={{e}^{x}}\cdot f'\left( {{e}^{x}} \right)+{{e}^{2x}}\cdot f''\left( {{e}^{x}} \right) …………………………(6)
Now, putting value x=1x=1 in equation (6) we get it as
ddx[ex.f(ex)]=ef(e)+e2f(e)\Rightarrow \dfrac{d}{dx}\left[ {{e}^{x}}.f'\left( {{e}^{x}} \right) \right]=e\cdot f'\left( e \right)+{{e}^{2}}\cdot f''\left( e \right)
On taking e common from the term, we get
ddx[ex.f(ex)]=e(f(e)+ef(e))\Rightarrow \dfrac{d}{dx}\left[ {{e}^{x}}.f'\left( {{e}^{x}} \right) \right]=e\left( f'\left( e \right)+e\cdot f''\left( e \right) \right)
Thus, we got the second order derivative as e(f(e)+ef(e))e\left( f'\left( e \right)+e\cdot f''\left( e \right) \right)
Hence, option (d) is the correct answer.

Note: Remember all the formulas and rules which are used in this type of sum. There are chances of mistakes when you consider f(ex)f\left( {{e}^{x}} \right) as f(x)f\left( x \right) . So, first derivative will be ddx(ex)=ex\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}} and again taking second derivative will give answer as ex{{e}^{x}} , if we use the property ddx(ex)=ex\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}} . So, this is not the correct method. You have to consider two different variables as shown in the given solution.