Question

Find the second derivative of the following function.
$y={{x}^{{{e}^{x}}}}$

Answer 1

Take ‘log’ to both the sides of the given equation. Now, use chain rule and multiplication rule of differentiation, which are given as Chain rule: $\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right).g'\left( x \right)$.
Multiplication Rule of differentiation: -
$\dfrac{d}{dx}\left( u\left( x \right).v\left( x \right) \right)=u\left( x \right)\dfrac{dv\left( x \right)}{dx}+v\left( x \right)\dfrac{du\left( x \right)}{dx}$

Complete step by step answer:
We have the given equation as
$y={{x}^{{{e}^{x}}}}$-(1)
Now, we can observe that y is an explicit function of type ${{\left( f\left( x \right) \right)}^{g\left( x \right)}}$.
Hence, we do not have a direct formula for the differentiation of equation (1).
Hence, for these kind of questions we need to take log to both the sides so that it gets easier to differentiate them in following way: -
$y={{x}^{{{e}^{x}}}}$
Taking log both sides: -

& \log y=\log {{x}^{{{e}^{x}}}}=\log {{\left( x \right)}^{{{e}^{x}}}} \\\ & \log y={{e}^{x}}\log x\left( \because \log {{a}^{n}}=n\log a \right) \\\ \end{aligned}$$ Now, we can differentiate the above function to both sides: - $$\dfrac{d}{dx}\left( \log y \right)=\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)-(2)$$ Let us differentiate LHS and RHS respectively: - Here, we have to apply chain rule as well i.e. differentiating all the functions involved one by one in following manner: - $$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right).{{g}^{'}}\left( x \right) \right)-(3)$$ For LHS part of equation (2) $$\begin{aligned} & \dfrac{d}{dx}\left( \log y \right)=\dfrac{1}{y}\dfrac{dy}{dx}\left( \because \dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x} \right) \\\ & \dfrac{d}{dx}\left( \log y \right)=\dfrac{1}{y}\dfrac{dy}{dx}-(4) \\\ \end{aligned}$$ Now, coming to the RHS part of equation (2) $$\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)$$ As we can see $$\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)$$contain two function $${{e}^{x}}$$and $$\log x$$in multiplication so we need to apply multiplication rule of differentiation as stated below: - Let us assume two function u(x) and v(x) in multiplication y=u(x)v(x) then $$\dfrac{dy}{dx}$$can be expressed as $$\dfrac{dy}{dx}=v\left( x \right)\dfrac{du\left( x \right)}{dx}+u\left( x \right)\dfrac{dv\left( x \right)}{dx}-(5)$$ Hence, applying multiplication rule in $$\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)$$where$$u\left( x \right)={{e}^{x}},v\left( x \right)=\log x$$ Hence, $$\begin{aligned} & \dfrac{d}{dx}\left( {{e}^{x}}\log x \right)={{e}^{x}}\dfrac{d}{dx}\left( \log x \right)+\log x\dfrac{d}{dx}\left( {{e}^{x}} \right) \\\ & \dfrac{d}{dx}\left( {{e}^{x}}\log x \right)=\dfrac{{{e}^{x}}}{x}+\log x.{{e}^{x}}\left( \because \dfrac{d}{dx}{{e}^{x}}={{e}^{x}},\dfrac{d}{dx}\log x=\dfrac{1}{x} \right) \\\ \end{aligned}$$ Therefore, for RHS part $$\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)=\dfrac{{{e}^{x}}}{x}+\log x.{{e}^{x}}-(6)$$ Now, coming to equation (2). It can be written as $$\begin{aligned} & \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \\\ & \dfrac{dy}{dx}=y\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)-(7) \\\ \end{aligned}$$ As we have to calculate the second derivative of the given function means we have to differentiate equation (7) again. Therefore, $$\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( y\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right) \right)$$ Now, we can observe that there are two functions in multiplication y and $$\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)$$in $$\dfrac{d}{dx}\left( y\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right) \right)$$. Therefore, we need to apply multiplication rule of differentiation as stated in equation (5) where $$u=y,v=\dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x$$ Let us simplify/calculate $$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$$ $$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( y\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right) \right) \\\ & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)\dfrac{dy}{dx}+y\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)-(8) \\\ \end{aligned}$$ Now, from equation (8) we see that we have to put values of $$\dfrac{dy}{dx}$$and $$\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)$$. As, we can put value of $$\dfrac{dy}{dx}$$from equation (7) $$\dfrac{dy}{dx}=y\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)$$ Hence, now we have to calculate $$\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)$$ $$\Rightarrow \dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{x} \right)+\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)-(9)$$ Let us assume $$A\left( x \right)=\dfrac{{{e}^{x}}}{x}$$and $$B\left( x \right)={{e}^{x}}\log x$$ $$\dfrac{dA}{dx}=\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{x} \right)$$ Here, we have to use division rule of differentiation as stated below: - $$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$$ Hence, from the function $$A\left( x \right)=\dfrac{{{e}^{x}}}{x}$$, we have $$u={{e}^{x}}$$and v=x Therefore, $$\begin{aligned} & \dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{x} \right)=\dfrac{x\dfrac{d}{dx}{{e}^{x}}-{{e}^{x}}\dfrac{d}{dx}\left( x \right)}{{{x}^{2}}} \\\ & \dfrac{dA}{dx}=\dfrac{x{{e}^{x}}-{{e}^{x}}}{{{x}^{2}}}\left( \because \dfrac{d}{dx}{{e}^{x}}={{e}^{x}},\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right) \\\ \end{aligned}$$ Now, for $$\dfrac{dB}{dx}$$or $$\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)$$we can apply multiplication rule of differentiation as stated in equation (5). $$\dfrac{d}{dx}uv=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$$ Hence, from the given function B(x) we can compare values of u and v $$\begin{aligned} & u\left( x \right)={{e}^{x}},v\left( x \right)=\log x \\\ & \dfrac{d}{dx}B\left( x \right)=\dfrac{d}{dx}\left( {{e}^{x}}\log x \right)={{e}^{x}}\dfrac{d}{dx}\log x+\log x\dfrac{d}{dx}{{e}^{x}} \\\ & \dfrac{dB}{dx}=\dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x\left( \because \dfrac{d}{dx}\log x=\dfrac{1}{x},\dfrac{d}{dx}{{e}^{x}}={{e}^{x}} \right) \\\ \end{aligned}$$ Hence, now put value in equation (9) of $$\dfrac{dA}{dx}$$and $$\dfrac{dB}{dx}$$ $$\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)=\dfrac{x{{e}^{x}}-{{e}^{x}}}{{{x}^{2}}}+\dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x$$ Now put the above value in equation (8) $$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)\left( y \right)\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)+y\left( \dfrac{x{{e}^{x}}-{{e}^{x}}}{{{x}^{2}}}+\dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right) \\\ & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=y\left[ {{\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)}^{2}}+\left( \dfrac{{{e}^{x}}}{x}+{{e}^{x}}\log x \right)+\dfrac{x{{e}^{x}}-{{e}^{x}}}{{{x}^{2}}} \right] \\\ \end{aligned}$$ Where $$y={{x}^{{{e}^{x}}}}$$ **Note:** We can go wrong by applying $$\dfrac{d}{dx}\left( {{x}^{n}} \right)$$in $$\dfrac{d}{dx}\left( {{x}^{{{e}^{x}}}} \right)$$by taking $${{e}^{x}}$$as a constant. But n is an integer but $${{e}^{x}}$$is a function. Hence, we can not apply $$\dfrac{d}{dx}\left( {{x}^{n}} \right)$$with $$\dfrac{d}{dx}\left( {{x}^{{{e}^{x}}}} \right)$$. Calculation is the beauty of these kinds of questions hence need to take care of writing the appropriate terms while differentiating. Students always need to take log on both sides while we have function of type $${{\left( f\left( x \right) \right)}^{g\left( x \right)}}$$.It can not be solved any other way. By taking a log to both sides, the question became flexible to solve.