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Exercise 12.2 · Q8

Q.Find the derivative of xn−anx−a\dfrac{x^n - a^n}{x - a} for some constant aa.

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This asks for the derivative of the function xn−anx−a\dfrac{x^n-a^n}{x-a} (with aa constant), so apply the Quotient Rule directly. The derivative is nxn−1(x−a)−(xn−an)(x−a)2\dfrac{nx^{n-1}(x-a)-(x^n-a^n)}{(x-a)^2}.

We must differentiate f(x)=xn−anx−af(x)=\dfrac{x^n-a^n}{x-a} with respect to xx, treating aa as a constant. This is a ratio of two functions, so the Quotient Rule is the tool.

Step 1 — Identify the parts and their derivatives.

Let u=xn−anu=x^n-a^n and v=x−av=x-a. Since aa is constant,

u′=ddx(xn−an)=nxn−1,v′=ddx(x−a)=1.u'=\frac{d}{dx}(x^n-a^n)=nx^{n-1},\qquad v'=\frac{d}{dx}(x-a)=1.

Step 2 — Apply the Quotient Rule (uv)′=u′v−uv′v2\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}:

f′(x)=nxn−1(x−a)−(xn−an)(1)(x−a)2=nxn−1(x−a)−(xn−an)(x−a)2.f'(x)=\frac{nx^{n-1}(x-a)-(x^n-a^n)(1)}{(x-a)^2}=\frac{nx^{n-1}(x-a)-(x^n-a^n)}{(x-a)^2}.

Step 3 — Expand the numerator (a neater equivalent form).

nxn−1(x−a)−(xn−an)=nxn−anxn−1−xn+an=(n−1)xn−naxn−1+an,nx^{n-1}(x-a)-(x^n-a^n)=nx^n-anx^{n-1}-x^n+a^n=(n-1)x^n-nax^{n-1}+a^n,

so

f′(x)=(n−1)xn−naxn−1+an(x−a)2.f'(x)=\frac{(n-1)x^n-nax^{n-1}+a^n}{(x-a)^2}. …

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