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Exercise 12.2 · Q10

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

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The derivative of cos⁡x\cos x from first principles is −sin⁡x-\sin x. This is found by applying the limit definition of the derivative, using the cosine difference identity, and evaluating two standard trigonometric limits.

The "first principle" method is the formal definition of a derivative. It asks: what happens to the function when we make an infinitesimally small change in xx? For any function f(x)f(x), the derivative at xx is defined as:

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

This is the slope of the tangent line, calculated as the limit of the slope of a secant line as the two points get infinitely close. For f(x)=cos⁡xf(x) = \cos x, we need to evaluate:

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

The challenge is that we cannot simply plug h=0h=0 — that gives 00\frac{0}{0}. We need to manipulate the expression using trigonometric identities until the limit becomes computable.

  1. Apply the cosine sum formula. The identity cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A \cos B - \sin A \sin B lets us expand cos⁡(x+h)\cos(x+h):

cos⁡(x+h)=cos⁡xcos⁡h−sin⁡xsin⁡h\cos(x+h) = \cos x \cos h - \sin x \sin h

Substituting into the limit:

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

  1. Factor and separate terms. Group the cos⁡x\cos x terms together:

lim⁡h→0cos⁡x(cos⁡h−1)−sin⁡xsin⁡hh\lim_{h \to 0} \frac{\cos x (\cos h - 1) - \sin x \sin h}{h}

Since the limit of a sum is the sum of the limits (provided each exists), we can split:

cos⁡x⋅lim⁡h→0cos⁡h−1h  −  sin⁡x⋅lim⁡h→0sin⁡hh\cos x \cdot \lim_{h \to 0} \frac{\cos h - 1}{h} \;-\; \sin x \cdot \lim_{h \to 0} \frac{\sin h}{h}

  1. Evaluate the two standard limits.

    These are the heart of the proof. You must know them:

    • lim⁡h→0sin⁡hh=1\displaystyle \lim_{h \to 0} \frac{\sin h}{h} = 1 (this is the fundamental trigonometric limit)
    • lim⁡h→0cos⁡h−1h=0\displaystyle \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 (this follows from the identity cos⁡h−1=−2sin⁡2(h/2)\cos h - 1 = -2\sin^2(h/2) and the previous limit)
    ›Proof

    Proof of lim⁡h→0cos⁡h−1h=0\displaystyle \lim_{h \to 0} \frac{\cos h - 1}{h} = 0:

    Use cos⁡h−1=−2sin⁡2(h2)\cos h - 1 = -2\sin^2\left(\frac{h}{2}\right). Then …

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