Answer

Verified

478.8k+ 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

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE