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,

& s=1+t{{e}^{s}} \\\ & \Rightarrow t{{e}^{s}}=s-1.........(ii) \\\ \end{aligned}$$ Substituting value from equation (ii) into equation (i), we get $$\begin{aligned} & \Rightarrow \dfrac{ds}{dt}\left( 1-(s-1) \right)={{e}^{s}} \\\ & \Rightarrow \dfrac{ds}{dt}\left( 2-s \right)={{e}^{s}} \\\ \end{aligned}$$ $$\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.