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Question: If \(y={{x}^{3}}+5\) and \(x\) changes from \(3\) to \(2.99\) , then the approximate change in \(y\)...

If y=x3+5y={{x}^{3}}+5 and xx changes from 33 to 2.992.99 , then the approximate change in yy is
1)2.72.7
2)0.27-0.27
3) 2727
4) None of these

Explanation

Solution

To solve this question use the concept of approximations of applications of derivatives. In a two variable equation we have one dependent variable and one independent variable. If we change the value of an independent variable then the value of the dependent variable will also change. To find change in dependent variable we use the concept of approximation.

Complete step-by-step solution:
Given equation is
y=x3+5y={{x}^{3}}+5……..(1)(1)
As we know,
ΔyΔx=dydx\dfrac{\Delta y}{\Delta x}=\dfrac{dy}{dx}
Δy\Delta y Denotes the change in yy
Δx\Delta x Denotes the change in xx
Δy=dydx.Δx\Rightarrow \Delta y=\dfrac{dy}{dx}.\Delta x ……….(2)(2)
Differentiating the given equation (1)(1)with respect toxx, we get
\Rightarrow dydx=3x2\dfrac{dy}{dx}=3{{x}^{2}}
Put this value in equation (2)(2)
Δy=3x2Δx\Rightarrow \Delta y=3{{x}^{2}}\Delta x ………(3)(3)
Since it is given that
Δx=2.993\Delta x=2.99-3 (( xx is changing from 33 to 2.992.99 ))
Δx=0.01\therefore \Delta x=-0.01
Put this value in equation (3)(3)

& \Rightarrow \Delta y=3{{x}^{2}}\times -0.01 \\\ & \Rightarrow \Delta y=3.{{(3)}^{2}}\times -0.01 \\\ & \Rightarrow \Delta y=27\times -0.01 \\\ & \Rightarrow \Delta y=-0.27 \\\ \end{aligned}$$ **Hence the correct answer is option $2$ .** **Note:** Derivative of a function tells us the value of change in function by changing the value of an independent variable. So $\dfrac{dy}{dx}$ tells us about the change in value of $y$ with respect to change in $x$ . So, the derivative of a function will tell us the change in value of a function. This derivative can also be used to find the maximum and minimum value of the function.