Question 36 of 70intermediate🔧 ApplyShort Answer2 marks

Find the derivative of xn+axn−1+a2xn−2+…+an−1x+anx^n + ax^{n-1} + a^2x^{n-2} + \ldots + a^{n-1}x + a^n for some fixed real number aa.

Correct Answer

The derivative is nxn−1+(n−1)axn−2+(n−2)a2xn−3+…+2an−2x+an−1nx^{n-1} + (n-1)ax^{n-2} + (n-2)a^2x^{n-3} + \ldots + 2a^{n-2}x + a^{n-1}.

Exercise: EXERCISE 12.2 | Q: 6 | (Page Page 32)
For More Understanding

Explanation

This question applies Theorem 7 (derivative of polynomials) from the textbook. The given expression is a polynomial in xx where the coefficients are powers of the fixed real number aa. Using the rule that derivative of xnx^n is nxn−1nx^{n-1} and applying it term by term with the appropriate coefficients, we get the derivative. The constant term ana^n has derivative zero.

Solution Steps

  1. Step 1: Identify the given expression as a polynomial: f(x)=xn+axn−1+a2xn−2+…+an−1x+anf(x) = x^n + ax^{n-1} + a^2x^{n-2} + \ldots + a^{n-1}x + a^n

  2. Step 2: Apply Theorem 7 which states that for a polynomial, we differentiate each term using the rule: derivative of xnx^n is nxn−1nx^{n-1}.

  3. Step 3: Differentiate each term: ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}, ddx(axn−1)=(n−1)axn−2\frac{d}{dx}(ax^{n-1}) = (n-1)ax^{n-2}, ddx(a2xn−2)=(n−2)a2xn−3\frac{d}{dx}(a^2x^{n-2}) = (n-2)a^2x^{n-3}, and so on.

  4. Step 4: The last two terms: ddx(an−1x)=an−1\frac{d}{dx}(a^{n-1}x) = a^{n-1} and ddx(an)=0\frac{d}{dx}(a^n) = 0 (constant term).

  5. Step 5: Combine all terms to get: nxn−1+(n−1)axn−2+(n−2)a2xn−3+…+2an−2x+an−1nx^{n-1} + (n-1)ax^{n-2} + (n-2)a^2x^{n-3} + \ldots + 2a^{n-2}x + a^{n-1}.

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