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Please solve rd sharma class 12 chapter 12 derivatives as a rate measure exercise fill in the blanks question 9 maths textbook solution

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Answer: \frac{15}{13}cm/s

Hint: Here we use the concept of acceleration and velocity

Given: V= 5x-\left ( x^{2}/6 \right )

Solution: So,

                \frac{dV}{dt}=5\times \frac{dx}{dt}-\frac{x}{3}\times \frac{dx}{dt}           

                \frac{dx}{dt}=\frac{\frac{dV}{dt}}{5-\left ( \frac{x}{3} \right )}

When x=2 and  \frac{dV}{dt}=5cm^{3}/sec

                \frac{dx}{dt}=\frac{5}{5-\frac{2}{3}}=\frac{15}{13}cm/s

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