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# If $y=x+{{e}^{x}}$ , then $\dfrac{{{d}^{2}}x}{d{{y}^{2}}}$ is equal to(a) ${{e}^{x}}$ (b) $-\dfrac{{{e}^{x}}}{{{\left( 1+{{e}^{x}} \right)}^{3}}}$ (c)$-\dfrac{{{e}^{x}}}{{{\left( 1+{{e}^{x}} \right)}^{2}}}$ (d) $\dfrac{{{e}^{x}}}{{{\left(1+{{e}^{x}} \right)}^{2}}}$ Verified
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Hint: The given expression should be derived with respect to $’y’$ and not $’x’$.

The given expression is
$y=x+{{e}^{x}}$
Now we will find the first order derivative of the given expression, so we will differentiate the given
expression with respect to $'y'$, we get
$\dfrac{d}{dy}(y)=\dfrac{d}{dy}\left( x+{{e}^{x}} \right)$
Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e.,
$\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$. Applying this formula in the above equation, we get
$\dfrac{d}{dy}(y)=\dfrac{d}{dy}\left( x \right)+\dfrac{d}{dy}\left( {{e}^{x}} \right)$
We know differentiation of an exponential function is, $\dfrac{d}{dx}\left( {{e}^{u}} \right)={{e}^{u}}.\dfrac{d}{dx}(u)$, so the above equation becomes,
$\dfrac{d}{dy}(y)=\dfrac{dx}{dy}+{{e}^{x}}\dfrac{d}{dy}\left( x \right)$
We know differentiation, $\dfrac{d}{dx}\left( x \right)=1$ , so the above equation becomes,
$1=\dfrac{dx}{dy}\left( 1+{{e}^{x}} \right)$
$\dfrac{dx}{dy}=\dfrac{1}{1+{{e}^{x}}}={{\left( 1+{{e}^{x}} \right)}^{-1}}........(i)$
Now we will find the second order derivative. For that we will again differentiate the above
expression with respect to $'y'$ , we get
$\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{d}{dy}{{\left( 1+{{e}^{x}} \right)}^{-1}}$
Now we know $\dfrac{d}{dx}({{u}^{n}})=n{{u}^{n-1}}\dfrac{d}{dx}(u)$ , applying this formula, the above equation becomes,
$\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=(-1){{\left( 1+{{e}^{x}} \right)}^{-1-1}}\dfrac{d}{dy}\left( 1+{{e}^{x}} \right)$
Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e.,
$\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$ . Applying this formula in the above equation, we get
$\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=(-1){{\left( 1+{{e}^{x}} \right)}^{-2}}\left[ \dfrac{d}{dy}\left( 1 \right)+\dfrac{d}{dy}\left( {{e}^{x}} \right) \right]$
We know differentiation of an exponential function is, $\dfrac{d}{dx}\left( {{e}^{u}} \right)={{e}^{u}}.\dfrac{d}{dx}(u)$, so the above equation becomes,
$\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=(-1){{\left( 1+{{e}^{x}} \right)}^{-2}}\left[ \dfrac{d}{dy}\left( 1 \right)+{{e}^{x}}\dfrac{d}{dy}\left( x \right) \right]$
We know differentiation of a constant is always a zero, so above equation can be written as,

$\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=(-1){{\left( 1+{{e}^{x}} \right)}^{-2}}\left[ 0+{{e}^{x}}\dfrac{dx}{dy} \right]$
Now substituting from equation (i), the above equation becomes,
\begin{align} & \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=(-1){{\left( 1+{{e}^{x}} \right)}^{-2}}\left[ {{e}^{x}}{{\left( 1+{{e}^{x}} \right)}^{-1}} \right] \\ & \Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-{{e}^{x}}}{{{\left( 1+{{e}^{x}} \right)}^{2+1}}} \\ & \Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-{{e}^{x}}}{{{\left( 1+{{e}^{x}} \right)}^{3}}} \\ \end{align}
Hence the correct answer is option (b).
Note: We know derivative of ${{e}^{x}}$ is ${{e}^{x}}$, but this is with respect to $x$, if the derivative is with respect to any other variable, then we cannot assume this, we should use the formula$\dfrac{d}{dx}\left( {{e}^{u}} \right)={{e}^{u}}.\dfrac{d}{dx}(u)$. Whenever derivating we should pay attention to the variable it is being derived with respect to. As in this problem see the simple expression student will derive with respect to $'x'$ instead of $'y'$ and will get an incorrect answer.

Last updated date: 21st Sep 2023
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