Question

If\[s=1+t{{e}^{s}}\], then find the value of \[\dfrac{{{d}^{2}}s}{d{{t}^{2}}}\]

Answer 1

Hint: Directly apply the derivative and apply necessary rules of differentiation. And the given expression should be derived with respect to $'t'$ and not $'x'$ or $'y'$.

Complete step by step answer:
 The given expression is
\[s=1+t{{e}^{s}}\]
Now we will find the first order derivative of the given expression, so we will differentiate the given expression with respect to $'t'$, we get
\[\dfrac{ds}{dt}=\dfrac{d}{dt}\left( 1+t{{e}^{s}} \right)\]
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
\[\dfrac{ds}{dt}=\dfrac{d}{dt}\left( 1 \right)+\dfrac{d}{dt}\left( t{{e}^{s}} \right)\]
We know the derivation of constant term is zero, so we get
\[\dfrac{ds}{dt}=0+\dfrac{d}{dt}\left( t{{e}^{s}} \right)\]
We know the product rule as, \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], applying this formula in the above equation, we get
\[\dfrac{ds}{dt}=t\dfrac{d}{dt}\left( {{e}^{s}} \right)+{{e}^{s}}\dfrac{d}{dt}\left( t \right)\]
We know differentiation of exponential is, $\dfrac{d}{dx}\left( {{e}^{u}} \right)={{e}^{u}}.\dfrac{d}{dx}(u)$, so the above equation becomes,
\[\dfrac{ds}{dt}=t{{e}^{s}}\dfrac{d}{dt}\left( s \right)+{{e}^{s}}(1)\]
Bringing the same terms on one side, we get
\[\dfrac{ds}{dt}(1-t{{e}^{s}})={{e}^{s}}..........(i)\]
Consider the given expression,
\[\begin{align}
  & s=1+t{{e}^{s}} \\
 & \Rightarrow t{{e}^{s}}=s-1.........(ii) \\
\end{align}\]
Substituting value from equation (ii) into equation (i), we get
\[\begin{align}
  & \Rightarrow \dfrac{ds}{dt}\left( 1-(s-1) \right)={{e}^{s}} \\
 & \Rightarrow \dfrac{ds}{dt}\left( 2-s \right)={{e}^{s}} \\
\end{align}\]
\[\Rightarrow \dfrac{ds}{dt}=\dfrac{{{e}^{s}}}{\left( 2-s \right)}.........(iii)\]
Now we will find the second order derivative. For that we will again differentiate the above expression with respect to $'t'$ , we get
\[\Rightarrow \dfrac{d}{dt}\left( \dfrac{ds}{dt} \right)=\dfrac{d}{dt}\left( \dfrac{{{e}^{s}}}{\left( 2-s \right)} \right)\]
Now we know the quotient rule, i.e., \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}\], applying this formula in the above equation, we get
\[\Rightarrow \dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\left( \dfrac{\left( 2-s \right)\dfrac{d}{dt}\left( {{e}^{s}} \right)-{{e}^{s}}\dfrac{d}{dt}\left( 2-s \right)}{{{\left( 2-s \right)}^{2}}} \right)\]
Applying the rules which are already mentioned above, we get
\[\Rightarrow \dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\left( \dfrac{\left( 2-s \right){{e}^{s}}\dfrac{ds}{dt}+{{e}^{s}}\dfrac{ds}{dt}}{{{\left( 2-s \right)}^{2}}} \right)\]
Taking out the common term, we get
\[\Rightarrow \dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\left( \dfrac{{{e}^{s}}\dfrac{ds}{dt}(2-s+1)}{{{\left( 2-s \right)}^{2}}} \right)\]
Now substituting value from equation (iii), we get
\[\Rightarrow \dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\left( \dfrac{{{e}^{s}}\left( \dfrac{{{e}^{s}}}{\left( 2-s \right)} \right)(3-s)}{{{\left( 2-s \right)}^{2}}} \right)\]
\[\Rightarrow \dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\left( \dfrac{{{\left( {{e}^{s}} \right)}^{2}}(3-s)}{{{\left( 2-s \right)}^{3}}} \right)\]
This is the required answer.

Note: We know derivate of \[{{e}^{x}}={{e}^{x}}\], 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 deriving we should pay attention to the variable it is being derived with respect to.
As in this problem see the simple expression students will derive with respect to $'x'$instead of $'y'$ and will get an incorrect answer.