Miscellaneous

Chapter 12 Class 11 Limits and Derivatives
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Misc 1 Find the derivative of the following functions from first principle: (iii) sin (x + 1) Let f (x) = sin (x + 1) 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 + 1) So, f (x + h) = sin ((x + h) + 1) Putting values fβ(x) = limβ¬(hβ0)β‘γsinβ‘γ(( π₯ + β ) + 1 ) βγ sinγβ‘γ (π₯ + 1)γ γ/hγ = limβ¬(hβ0)β‘γ(π ππ( π₯ + 1 + β) βγ sinγβ‘γ ( π₯ + 1 )γ)/hγ Using sin A β sin B = 2 cos ((π΄ + π΅)/2) sin ((π΄ β π΅)/2) = limβ¬(hβ0)β‘γ(2 cosβ‘(( ( π₯ + 1 + β ) + ( π₯ + 1))/2).sinβ‘((( π₯ + 1 + β ) β ( π₯ + 1 ))/2))/hγ = limβ¬(hβ0)β‘γ(2 cosβ‘γ(( 2(π₯ + 1 )+ β))/2γ . γsin γβ‘γβ/2γ)/hγ = limβ¬(hβ0)β‘γ(cosβ‘γ(( 2(π₯ + 1 )+ β))/2γ . γsin γβ‘γβ/2γ)/(β/2)γ = limβ¬(hβ0)β‘γγcos γβ‘γ(( 2(π₯ + 1 )+ β))/2γ . sinβ‘γ β/2γ/(( β)/2)γ = limβ¬(hβ0)β‘γcos γβ‘γ(( 2(π₯ + 1 )+ β))/2γ .(π₯π’π¦)β¬(π‘βπ) πππβ‘γ π/πγ/(( π)/π) = limβ¬(hβ0)β‘γcos γβ‘γ(( 2(π₯ + 1 ) + β))/2γ .π = limβ¬(hβ0)β‘γcos γβ‘γ(( 2(π₯ + 1 )+ β))/2γ Putting h = 0 =γ cos γβ‘γ(( 2(π₯ + 1 ) + 0))/2γ = cos ( x + 1) Hence fβ (x) = cos (x + 1) Using (πππ)β¬(π₯β0)β‘γ π ππβ‘π₯/π₯γ=1 Replacing x by β/2 β (πππ)β¬(ββ0) π ππβ‘γ β/2γ/(( β)/2) = 1