Examples

Chapter 12 Class 11 Limits and Derivatives
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Transcript

Example 20 Find the derivative of f(x) from the first principle, where f(x) is (i) sin x + cos x Given f (x) = sin x + cos x We need to find Derivative of f(x) We know that fβ(x) = limβ¬(hβ0) πβ‘γ(π₯ + β) β π(π₯)γ/β Here, f (x) = sin x + cos x f (x + h) = sin (x + h) + cos (x + h) Putting values fβ(x) = limβ¬(hβ0)β‘γ(sinβ‘γ(π₯ + β)γ + cosβ‘(π₯ + β) β (sinβ‘π₯ + cosβ‘γπ₯)γ)/βγ Using sin (A + B) = sin A cos B + cos A sin B & cos (A + B) = cos A cos B β sin A sin B = limβ¬(hβ0)β‘γsinβ‘γπ₯ cosβ‘γβ +γ cosγβ‘γπ₯ sinβ‘γβ + cosβ‘γπ₯ cosβ‘γβ β sinβ‘γπ₯ γ sinγβ‘γβ βγ sinγβ‘γπ₯ βγ cosγβ‘π₯ γ γ γ γ γ γ γ γ γ/hγ = limβ¬(hβ0)β‘γcosβ‘γπ₯ sinβ‘γβ βγ sinγβ‘γπ₯ sinβ‘γβ + sinβ‘γπ₯ cosβ‘γβ β sinβ‘γπ₯ +γ cosγβ‘γπ₯ cosβ‘γβ βγ cosγβ‘π₯ γ γ γ γ γ γ γ γ γ/hγ = limβ¬(hβ0)β‘γsinβ‘γβ γ(cosγβ‘γπ₯ β sinβ‘γπ₯) + sinβ‘γπ₯ (cosβ‘γβ β 1) + cosβ‘π₯ (cosβ‘γβ β 1)γ γ γ γ γ γ/hγ = limβ¬(hβ0)β‘(sinβ‘γβ γ(cosγβ‘γπ₯ β sinβ‘γπ₯)γ γ γ/h+sinβ‘γπ₯ (cosβ‘γβ β 1)γ γ/h+cosβ‘γπ₯ (cosβ‘γβ β 1)γ γ/β)" " = limβ¬(hβ0)β‘γsinβ‘γβ γ(cosγβ‘γπ₯ βγ sinγβ‘γπ₯)γ γ γ/h+limβ¬(hβ0) sinβ‘γπ₯ (cosβ‘γβ β 1)γ γ/h+limβ¬(hβ0) cosβ‘γπ₯ (cosβ‘γβ β 1)γ γ/βγ = limβ¬(hβ0)β‘γ"(cos x β sin x)" sinβ‘β/β+limβ¬(hβ0) "(β sin x)" ((1 β cosβ‘γβ)γ)/βγ+limβ¬(hβ0) "(β cos x)" ((1 β cosβ‘γβ)γ)/β = "(cos x β sin x)" (π₯π’π¦)β¬(π‘βπ)β‘γπ¬π’π§β‘π/πβsinβ‘γπ₯ (π₯π’π¦)β¬(π‘βπ) γ ((π β πππβ‘γπ)γ)/πβcosβ‘π₯ (π₯π’π¦)β¬(π‘βπ) ((π β πππβ‘γπ)γ)/πγUsing (πππ)β¬(ββ0) π ππβ‘β/β = 1 & (πππ)β¬(ββ0) γ(1 β πππ γβ‘γβ)γ/β = 0 Using (πππ)β¬(ββ0) π ππβ‘β/β = 1 & (πππ)β¬(ββ0) γ(1 β πππ γβ‘γβ)γ/β = 0