Question 55 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): cosec⁡xcot⁡x\cosec x \cot x

Correct Answer

−cosec⁡x(cosec⁡2x+cot⁡2x)-\cosec x(\cosec^2 x + \cot^2 x)

Exercise: Miscellaneous Exercise on Chapter 12 | Q: 15 | (Page Page 37)
For More Understanding

Explanation

Using product rule on cosec⁡xcot⁡x\cosec x \cot x: First, ddx(cosec⁡x)=−cosec⁡xcot⁡x\frac{d}{dx}(\cosec x) = -\cosec x \cot x (from quotient rule on 1sin⁡x\frac{1}{\sin x}). Second, ddx(cot⁡x)=−cosec⁡2x\frac{d}{dx}(\cot x) = -\cosec^2 x (from context). Applying product rule: ddx(cosec⁡xcot⁡x)=cosec⁡x⋅(−cosec⁡2x)+cot⁡x⋅(−cosec⁡xcot⁡x)=−cosec⁡3x−cosec⁡xcot⁡2x=−cosec⁡x(cosec⁡2x+cot⁡2x)\frac{d}{dx}(\cosec x \cot x) = \cosec x \cdot (-\cosec^2 x) + \cot x \cdot (-\cosec x \cot x) = -\cosec^3 x - \cosec x \cot^2 x = -\cosec x(\cosec^2 x + \cot^2 x).

Solution Steps

  1. Step 1: Apply product rule: ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv' where u=cosec⁡xu = \cosec x and v=cot⁡xv = \cot x

  2. Step 2: Find ddx(cosec⁡x)=−cosec⁡xcot⁡x\frac{d}{dx}(\cosec x) = -\cosec x \cot x using quotient rule on 1sin⁡x\frac{1}{\sin x}

  3. Step 3: Use ddx(cot⁡x)=−cosec⁡2x\frac{d}{dx}(\cot x) = -\cosec^2 x from context

  4. Step 4: Apply product rule: cosec⁡x⋅(−cosec⁡2x)+cot⁡x⋅(−cosec⁡xcot⁡x)=−cosec⁡3x−cosec⁡xcot⁡2x\cosec x \cdot (-\cosec^2 x) + \cot x \cdot (-\cosec x \cot x) = -\cosec^3 x - \cosec x \cot^2 x

  5. Step 5: Factor to get −cosec⁡x(cosec⁡2x+cot⁡2x)-\cosec x(\cosec^2 x + \cot^2 x)

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