Question 40 of 70intermediate🔧 ApplyLong Answer3 marks

Find the derivative of cos⁡x\cos x from first principle.

Correct Answer

Let f(x)=cos⁡xf(x) = \cos x. By the first principle, the derivative is given by:

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0cos⁡(x+h)−cos⁡xhf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}

Using the trigonometric identity cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right):

=lim⁡h→0−2sin⁡(x+h2)sin⁡(h2)h= \lim_{h \to 0} \frac{-2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} =lim⁡h→0[−sin⁡(x+h2)⋅sin⁡(h2)h2]= \lim_{h \to 0} \left[-\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]

As h→0h \to 0, we have sin⁡(h2)h2→1\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \to 1 and sin⁡(x+h2)→sin⁡x\sin\left(x + \frac{h}{2}\right) \to \sin x.

Therefore, ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x

∴ ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = \mathbf{-\sin x}

Exercise: EXERCISE 12.2 | Q: 10 | (Page Page 32)
For More Understanding

Explanation

This question follows the same pattern as Example 16 in the textbook, which computes the derivative of sin⁡x\sin x from first principle. The key steps involve:

  1. Applying the definition of derivative from first principle.
  2. Using the trigonometric identity for cos⁡A−cos⁡B\cos A - \cos B to simplify the numerator.
  3. Applying the standard limit result that lim⁡h→0[sin⁡(h/2)h/2]=1\lim_{h \to 0}\left[\frac{\sin(h/2)}{h/2}\right] = 1.

The final result is −sin⁡x-\sin x, which matches the known derivative of cos⁡x\cos x.

Solution Steps

  1. Step 1: Apply the first principle definition:

    f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}

  2. Step 2: Substitute f(x)=cos⁡xf(x) = \cos x:

    lim⁡h→0cos⁡(x+h)−cos⁡xh\lim_{h \to 0}\frac{\cos(x+h) - \cos x}{h}

  3. Step 3: Use identity cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) to expand numerator:

    lim⁡h→0−2sin⁡((x+h)+x2)sin⁡((x+h)−x2)h\lim_{h \to 0}\frac{-2\sin\left(\frac{(x+h)+x}{2}\right)\sin\left(\frac{(x+h)-x}{2}\right)}{h}

  4. Step 4: Simplify to get:

    −sin⁡(x+h2)⋅sin⁡(h2)h2-\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}

  5. Step 5: Apply limits: sin⁡(h2)h2→1\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \to 1 and sin⁡(x+h2)→sin⁡x\sin\left(x + \frac{h}{2}\right) \to \sin x as h→0h \to 0.

  6. Step 6: Final answer:

    d(cos⁡x)dx=−sin⁡x\frac{d(\cos x)}{dx} = \mathbf{-\sin x}

    ∴ d(cos⁡x)dx=−sin⁡x\frac{d(\cos x)}{dx} = -\sin x

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