Question

If \[{\text{x = f(t)}}\] and \[{\text{y = g(t)}}\] are differentiable function of \[{\text{t}}\] then \[\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}\] is
A. \[ \dfrac{{{\text{f '(t)}}{\text{.g ''(t) - g '(t)}}{\text{.f ''(t)}}}}{{{{\left( {{\text{f '(t)}}} \right)}^{\text{3}}}}}\]
B. \[\dfrac{{{{\text{f}}^{\text{'}}}{\text{(t)}}{\text{.}}{{\text{g}}^{{\text{''}}}}{\text{(t) - }}{{\text{g}}^{\text{'}}}{\text{(t)}}{\text{.}}{{\text{f}}^{''}}{\text{(t)}}}}{{{{\left( {{{\text{f}}^{\text{'}}}{\text{(t)}}} \right)}^2}}}\]
C. \[\dfrac{{{\text{ }}{{\text{g}}^{\text{'}}}{\text{(t)}}{\text{.}}{{\text{f}}^{''}}{\text{(t) - }}{{\text{f}}^{\text{'}}}{\text{(t)}}{\text{.}}{{\text{g}}^{{\text{''}}}}{\text{(t) }}}}{{{{\left( {{{\text{f}}^{\text{'}}}{\text{(t)}}} \right)}^3}}}\]
D. \[\dfrac{{{\text{ }}{{\text{g}}^{\text{'}}}{\text{(t)}}{\text{.}}{{\text{f}}^{{\text{''}}}}{\text{(t) + }}{{\text{f}}^{\text{'}}}{\text{(t)}}{\text{.}}{{\text{g}}^{{\text{''}}}}{\text{(t) }}}}{{{{\left( {{{\text{f}}^{\text{'}}}{\text{(t)}}} \right)}^3}}}\]

Answer 1

Hint: In this question we are going to find the value for the given term
And here we are given with values of some variables
Using these we are going to solve the problem
Firstly, we need to differentiate the given values
After differentiation we need to find the value of given term
For that we are going divide the both the values we get
Then we want differentiate the values again that we got and we want to write them according to the chain rule
Then we need to apply the values that we have to the terms and we need to simplify them
After Simplification we will get the result which is required.

Formula used: The formula which is required to solve the problem was
\[\dfrac{{\text{u}}}{{\text{v}}}{\text{ = }}\dfrac{{{\text{vu' - uv'}}}}{{{{\text{v}}^{\text{2}}}}}\]

Complete step-by-step answer:
In the question they have given that \[{\text{x = f(t)}}\] and \[{\text{y = g(t)}}\]
While differentiating \[{\text{x}}\] with respect to \[{\text{t}}\]we get
\[\dfrac{{{\text{dx}}}}{{{\text{dt}}}}{\text{ = }}{{\text{f}}^{\text{'}}}{\text{(t) = }}{{\text{f}}^{\text{'}}}\]
Now we are going to differentiate \[{\text{y}}\] with respect to \[{\text{t}}\]
$\Rightarrow$\[\dfrac{{{\text{dy}}}}{{{\text{dt}}}}{\text{ = }}{{\text{g}}^{\text{'}}}{\text{(t) = }}{{\text{g}}^{\text{'}}}\]
Now, we are going to find the value of \[\dfrac{{{\text{dy}}}}{{{\text{dx}}}}\]
$\Rightarrow$\[\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\dfrac{{{\text{dy}}}}{{{\text{dt}}}}}}{{\dfrac{{{\text{dx}}}}{{{\text{dt}}}}}}\]
$\Rightarrow$\[\dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{{{\text{g}}^{\text{'}}}}}{{{{\text{f}}^{\text{'}}}}}\]
Again, we are going to differentiate the values that we got,
After differentiation we will get
$\Rightarrow$\[\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{f}}^{\text{'}}}\dfrac{{{\text{d(}}{{\text{g}}^{\text{'}}}{\text{)}}}}{{{\text{dx}}}}{\text{ - }}{{\text{g}}^{\text{'}}}\dfrac{{{\text{d(}}{{\text{f}}^{\text{'}}}{\text{)}}}}{{{\text{dx}}}}}}{{{{{\text{(}}{{\text{f}}^{\text{'}}}{\text{)}}}^{\text{2}}}}}{\text{ }}\]
Since \[\dfrac{{\text{u}}}{{\text{v}}}{\text{ = }}\dfrac{{{\text{vu' -uv'}}}}{{{{\text{v}}^{\text{2}}}}}\] formula
By the chain rule,
$\Rightarrow$\[\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{f}}^{\text{'}}}\dfrac{{{\text{d(}}{{\text{g}}^{\text{'}}}{\text{)}}}}{{{\text{dt}}}}\dfrac{{{\text{dt}}}}{{{\text{dx}}}}{\text{ - }}{{\text{g}}^{\text{'}}}\dfrac{{{\text{d(}}{{\text{f}}^{\text{'}}}{\text{)}}}}{{{\text{dt}}}}\dfrac{{{\text{dt}}}}{{{\text{dx}}}}}}{{{{{\text{(}}{{\text{f}}^{\text{'}}}{\text{)}}}^{\text{2}}}}}{\text{ }}\]
We already know that \[\dfrac{{{\text{dx}}}}{{{\text{dt}}}}{\text{ = }}{{\text{f}}^{\text{'}}}\]
So, while we reciprocal these we get that \[\dfrac{{{\text{dt}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{1}{{{{\text{f}}^{\text{'}}}}}\]
By applying the value, we get on the above we will have
\[ = \dfrac{{{{\text{f}}^{\text{'}}}{{\text{g}}^{{\text{''}}}}{\text{ \times }}\dfrac{{\text{1}}}{{{{\text{f}}^{\text{'}}}}}{\text{ - }}{{\text{g}}^{\text{'}}}{{\text{f}}^{{\text{''}}}}{\text{ \times }}\dfrac{{\text{1}}}{{{{\text{f}}^{\text{'}}}}}}}{{{{{\text{(}}{{\text{f}}^{\text{'}}}{\text{)}}}^{\text{2}}}}}\]
By taking out the term \[\dfrac{1}{{{{\text{f}}^{\text{'}}}}}\]which is common, we get that
\[ = \dfrac{{\left( {{{\text{f}}^{\text{'}}}{{\text{g}}^{{\text{''}}}}{\text{ - }}{{\text{g}}^{\text{'}}}{{\text{f}}^{{\text{''}}}}} \right)\dfrac{{\text{1}}}{{{{\text{f}}^{\text{'}}}}}}}{{{{{\text{(}}{{\text{f}}^{\text{'}}}{\text{)}}}^{\text{2}}}}}\]
Here we are Bringing the term \[\dfrac{{\text{1}}}{{{{\text{f}}^{\text{'}}}}}\] to the denominator part we will get that
\[ = \dfrac{{{{\text{f}}^{\text{'}}}{{\text{g}}^{{\text{''}}}} - {{\text{g}}^{\text{'}}}{{\text{f}}^{{\text{''}}}}}}{{{{{\text{(}}{{\text{f}}^{\text{'}}}{\text{)}}}^3}}}\]
Therefore, we finally got the value of \[\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}\] that is \[\dfrac{{{{\text{f}}^{\text{'}}}{{\text{g}}^{{\text{''}}}} - {{\text{g}}^{\text{'}}}{{\text{f}}^{{\text{''}}}}}}{{{{{\text{(}}{{\text{f}}^{\text{'}}}{\text{)}}}^3}}}\]
\[\dfrac{{{{\text{d}}^{\text{2}}}{\text{y}}}}{{{\text{d}}{{\text{x}}^{\text{2}}}}}\]\[ = \dfrac{{{\text{f '(t)}}{\text{.g ''(t) - g '(t)}}{\text{.f ''(t)}}}}{{{{\left( {{\text{f '(t)}}} \right)}^{\text{3}}}}}\]

Option A is the correct answer.

Note: A differentiable function of one real variable is a function whose derivative exists at each point in its domain. Chain rule is a basic concept of differentiation and it’s very important. The chain rule tells us how to find the derivative of a composite function.