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Exercise 8.7 · Q2

Q.Let the function y=x2y = x^2 measure the area of a metallic square of side xx. If at any given time the side of the square is aa, and we heat the square, uniformly increasing the side, what is the tendency of change of the area at that moment?

Andaman Nicobar CbseNCERTSubjective· 2mImportance★★★★★
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The "tendency of change" of area with respect to side length is the derivative dydx\dfrac{dy}{dx}, evaluated at x=ax=a.

For y=f(x)y=f(x), the instantaneous rate of change of yy with respect to xx at x=x0x=x_0 is the derivative evaluated there:

dydx∣x=x0,with ddx(xn)=nxn−1\left.\frac{dy}{dx}\right|_{x=x_0}, \qquad \text{with } \frac{d}{dx}(x^n) = nx^{n-1}

  1. Write the given relation. The area of a metallic square of side xx is

y=x2y = x^2

  1. Differentiate yy with respect to xx using the power rule ddx(xn)=nxn−1\frac{d}{dx}(x^n)=nx^{n-1} with n=2n=2:

dydx=2x2−1=2x\frac{dy}{dx} = 2x^{2-1} = 2x

This derivative represents the rate at which the area changes per unit increase in the side length, at a general side xx.

  1. Evaluate at the given instant, x=ax=a. dydx∣x=a=2a\left.\frac{dy}{dx}\right|_{x=a} = 2a …

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