Question 64 of 70intermediate🔧 ApplyShort Answer1 mark

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (ax2+sin⁡x)(p+qcos⁡x)(ax^2 + \sin x)(p + q \cos x)

Correct Answer

(2ax+cos⁡x)(p+qcos⁡x)−qsin⁡x(ax2+sin⁡x)(2ax + \cos x)(p + q \cos x) - q \sin x(ax^2 + \sin x)

Exercise: Miscellaneous Exercise on Chapter 12 | Q: 24 | (Page Page 37)
For More Understanding

Explanation

This question is from the Miscellaneous Exercise on Chapter 12 (Limits and Derivatives). The student needs to apply the product rule for differentiation since the function is a product of two expressions: (ax2+sin⁡x)(ax^2 + \sin x) and (p+qcos⁡x)(p + q \cos x). The product rule states that if y=uvy = uv, then y′=u′v+uv′y' = u'v + uv'.

Solution Steps

  1. Step 1: Let u=ax2+sin⁡xu = ax^2 + \sin x and v=p+qcos⁡xv = p + q \cos x.

  2. Step 2: Find u′=ddx(ax2+sin⁡x)=2ax+cos⁡xu' = \frac{d}{dx}(ax^2 + \sin x) = 2ax + \cos x.

  3. Step 3: Find v′=ddx(p+qcos⁡x)=0−qsin⁡x=−qsin⁡xv' = \frac{d}{dx}(p + q \cos x) = 0 - q \sin x = -q \sin x.

  4. Step 4: Apply the product rule: ddx(uv)=u′v+uv′=(2ax+cos⁡x)(p+qcos⁡x)+(ax2+sin⁡x)(−qsin⁡x)\frac{d}{dx}(uv) = u'v + uv' = (2ax + \cos x)(p + q \cos x) + (ax^2 + \sin x)(-q \sin x).

  5. Step 5: Final answer: (2ax+cos⁡x)(p+qcos⁡x)−qsin⁡x(ax2+sin⁡x)(2ax + \cos x)(p + q \cos x) - q \sin x(ax^2 + \sin x).

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