Answer
456.6k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)