Question 29 of 71intermediate🔧 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+a)cos⁡x\frac{\sin(x + a)}{\cos x}

Correct Answer

cos⁡asec⁡2x\cos a \sec^2 x

Exercise: Miscellaneous Exercise on Chapter 12 | Q: 21 | (Chapter: Page 37)
For More Understanding

Explanation

This question asks for the derivative of a quotient of trigonometric functions. Using the quotient rule and chain rule, we differentiate the numerator and denominator separately, then apply the standard formula for derivatives of quotients.

Solution Steps

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

  2. Step 2: Derivative of numerator: ddx[sin⁡(x+a)]=cos⁡(x+a)⋅1=cos⁡(x+a)\frac{d}{dx}[\sin(x + a)] = \cos(x + a) \cdot 1 = \cos(x + a) (using chain rule)

  3. Step 3: Derivative of denominator: ddx[cos⁡x]=−sin⁡x\frac{d}{dx}[\cos x] = -\sin x

  4. Step 4: Apply quotient rule: f′(x)=cos⁡(x+a)⋅cos⁡x−sin⁡(x+a)⋅(−sin⁡x)cos⁡2x=cos⁡(x+a)cos⁡x+sin⁡(x+a)sin⁡xcos⁡2xf'(x) = \frac{\cos(x + a) \cdot \cos x - \sin(x + a) \cdot (-\sin x)}{\cos^2 x} = \frac{\cos(x + a)\cos x + \sin(x + a)\sin x}{\cos^2 x}

  5. Step 5: Using identity cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B with A=x+aA = x + a and B=xB = x: numerator =cos⁡a= \cos a. Thus f′(x)=cos⁡acos⁡2x=cos⁡asec⁡2xf'(x) = \frac{\cos a}{\cos^2 x} = \cos a \sec^2 x

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