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

If 2t=v22t = {v^2}, then dvdt=\dfrac{{dv}}{{dt}} =
A. 00
B. 14\dfrac{1}{4}
C. 12\dfrac{1}{2}
D. 1v\dfrac{1}{v}

Explanation

Solution

Here, the given question. We have to find the derivative or differentiated term of the function. For this, first consider the given, then differentiate vv with respect to xx 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(x)y = f\left( x \right) be a function of. Then, the rate of change of “y” per unit change in “x” is given by dydx\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,
dydx=dydududx\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}
Consider the given function
2t=v2\Rightarrow \,\,\,\,2t = {v^2}---------- (1)
Here, vv is a dependent variable and tt is an independent variable.

Now we have to differentiate this function with respect to tt
ddt(2t)=ddt(v2)\Rightarrow \,\,\,\,\dfrac{d}{{dt}}\left( {2t} \right) = \dfrac{d}{{dt}}\left( {{v^2}} \right)
2dtdt=ddt(v2)\Rightarrow \,\,\,\,2\dfrac{{dt}}{{dt}} = \dfrac{d}{{dt}}\left( {{v^2}} \right)
Using the standard differentiated formula ddx(xn)=nxn1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}, then equation (2) becomes
21=2vdvdt\Rightarrow \,\,\,\,2 \cdot 1 = 2v \cdot \dfrac{{dv}}{{dt}}
2=2vdvdt\Rightarrow \,\,\,\,2 = 2v \cdot \dfrac{{dv}}{{dt}}
Divide both side by 2v2v, then
22v=dvdt\Rightarrow \,\,\,\,\dfrac{2}{{2v}} = \dfrac{{dv}}{{dt}}
On cancelling the like terms 22 in both numerator and denominator of LHS, then we get
1v=dvdt\Rightarrow \,\,\,\,\dfrac{1}{v} = \dfrac{{dv}}{{dt}}
dvdt=1v\therefore \,\,\,\,\dfrac{{dv}}{{dt}} = \dfrac{1}{v}
Hence, the required differentiated value dvdt=1v\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.