Question 45 of 70intermediate🔧 ApplyShort Answer1 mark

Find the derivative of the following functions (it is to be understood that a,b,c,d,p,q,ra, b, c, d, p, q, r and ss are fixed non-zero constants and mm and nn are integers): (ax+b)(cx+d)2(ax + b)(cx + d)^2

Correct Answer

(cx+d)[a(cx+d)+2c(ax+b)](cx + d)[a(cx + d) + 2c(ax + b)] or equivalently a(cx+d)2+2c(ax+b)(cx+d)a(cx + d)^2 + 2c(ax + b)(cx + d)

Exercise: Miscellaneous Exercise on Chapter 12 | Q: 4 | (Page Page 37)
For More Understanding

Explanation

This question from the Miscellaneous Exercise on Chapter 12 tests the application of the product rule combined with the chain rule. The function is a product of (ax+b)(ax + b) and (cx+d)2(cx + d)^2. Students must apply the product rule: derivative of product equals first function times derivative of second plus derivative of first times second function. The chain rule is needed for (cx+d)2(cx + d)^2.

Solution Steps

  1. Step 1: Let u=ax+bu = ax + b and v=(cx+d)2v = (cx + d)^2. Then u′=au' = a and v′=2(cx+d)⋅c=2c(cx+d)v' = 2(cx + d) \cdot c = 2c(cx + d) by chain rule.

  2. Step 2: Apply product rule: ddx(uv)=u′v+uv′=a(cx+d)2+(ax+b)⋅2c(cx+d)\frac{d}{dx}(uv) = u'v + uv' = a(cx + d)^2 + (ax + b) \cdot 2c(cx + d).

  3. Step 3: Factor out (cx+d)(cx + d): =(cx+d)[a(cx+d)+2c(ax+b)]= (cx + d)[a(cx + d) + 2c(ax + b)].

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