Question 57 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): sin⁡x+cos⁡xsin⁡x−cos⁡x\frac{\sin x + \cos x}{\sin x - \cos x}

Correct Answer

−2(sin⁡x−cos⁡x)2\frac{-2}{(\sin x - \cos x)^2}

Exercise: Miscellaneous Exercise on Chapter 12 | Q: 17 | (Page Page 37)
For More Understanding

Explanation

This question requires finding the derivative of a quotient of trigonometric functions. The student needs to apply the quotient rule along with the derivatives of sin x and cos x. From the context, we know that the derivative of sin x is cos x (Example 16) and the derivative of cos x is -sin x. The quotient rule is demonstrated in the context for similar problems.

Solution Steps

  1. Step 1: Let f(x)=sin⁡x+cos⁡xsin⁡x−cos⁡xf(x) = \frac{\sin x + \cos x}{\sin x - \cos x}. Apply the quotient rule: f′(x)=(sin⁡x+cos⁡x)′(sin⁡x−cos⁡x)−(sin⁡x+cos⁡x)(sin⁡x−cos⁡x)′(sin⁡x−cos⁡x)2f'(x) = \frac{(\sin x + \cos x)'(\sin x - \cos x) - (\sin x + \cos x)(\sin x - \cos x)'}{(\sin x - \cos x)^2}

  2. Step 2: Find derivatives: (sin⁡x+cos⁡x)′=cos⁡x−sin⁡x(\sin x + \cos x)' = \cos x - \sin x and (sin⁡x−cos⁡x)′=cos⁡x+sin⁡x(\sin x - \cos x)' = \cos x + \sin x

  3. Step 3: Substitute: f′(x)=(cos⁡x−sin⁡x)(sin⁡x−cos⁡x)−(sin⁡x+cos⁡x)(cos⁡x+sin⁡x)(sin⁡x−cos⁡x)2f'(x) = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}

  4. Step 4: Expand numerator: (cos⁡x−sin⁡x)(sin⁡x−cos⁡x)=2sin⁡xcos⁡x−sin⁡2x−cos⁡2x(\cos x - \sin x)(\sin x - \cos x) = 2\sin x \cos x - \sin^2 x - \cos^2 x and (sin⁡x+cos⁡x)(cos⁡x+sin⁡x)=2sin⁡xcos⁡x+sin⁡2x+cos⁡2x(\sin x + \cos x)(\cos x + \sin x) = 2\sin x \cos x + \sin^2 x + \cos^2 x

  5. Step 5: Subtract: Numerator =−2sin⁡2x−2cos⁡2x=−2(sin⁡2x+cos⁡2x)=−2= -2\sin^2 x - 2\cos^2 x = -2(\sin^2 x + \cos^2 x) = -2

  6. Step 6: Final answer: f′(x)=−2(sin⁡x−cos⁡x)2f'(x) = \frac{-2}{(\sin x - \cos x)^2}

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