Question

If \[2t = {v^2}\], then \[\dfrac{{dv}}{{dt}} = \]
A. \[0\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{1}{2}\]
D. \[\dfrac{1}{v}\]

Answer 1

Hint: Here, the given question. We have to find the derivative or differentiated term of the function. For this, first consider the given, then differentiate \[v\] with respect to \[x\] by using a standard differentiation formula and use chain rule for differentiation then on further simplification we get the required differentiation value.

Complete step by step answer:
Differentiation can be defined as a derivative of a function with respect to an independent variable. Otherwise, the differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists. Let \[y = f\left( x \right)\] be a function of. Then, the rate of change of “y” per unit change in “x” is given by \[\dfrac{{dy}}{{dx}}\].
The Chain Rule is a formula for computing the derivative of the composition of two or more functions. The chain rule expressed as,
\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}\]
Consider the given function
\[ \Rightarrow \,\,\,\,2t = {v^2}\]---------- (1)
Here, \[v\] is a dependent variable and \[t\] is an independent variable.

Now we have to differentiate this function with respect to \[t\]
\[ \Rightarrow \,\,\,\,\dfrac{d}{{dt}}\left( {2t} \right) = \dfrac{d}{{dt}}\left( {{v^2}} \right)\]
\[ \Rightarrow \,\,\,\,2\dfrac{{dt}}{{dt}} = \dfrac{d}{{dt}}\left( {{v^2}} \right)\]
Using the standard differentiated formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], then equation (2) becomes
\[ \Rightarrow \,\,\,\,2 \cdot 1 = 2v \cdot \dfrac{{dv}}{{dt}}\]
\[ \Rightarrow \,\,\,\,2 = 2v \cdot \dfrac{{dv}}{{dt}}\]
Divide both side by \[2v\], then
\[ \Rightarrow \,\,\,\,\dfrac{2}{{2v}} = \dfrac{{dv}}{{dt}}\]
On cancelling the like terms \[2\] in both numerator and denominator of LHS, then we get
\[ \Rightarrow \,\,\,\,\dfrac{1}{v} = \dfrac{{dv}}{{dt}}\]
\[\therefore \,\,\,\,\dfrac{{dv}}{{dt}} = \dfrac{1}{v}\]
Hence, the required differentiated value \[\dfrac{{dv}}{{dt}} = \dfrac{1}{v}\].

Therefore, option D is the correct answer.

Note: When differentiating the function or term the student must recognize the dependent and independent variable then differentiate the dependent variable with respect to independent variable and should remember the standard differentiation formulas.