Last updated at Aug. 28, 2021 by
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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¬(hβ0) πβ‘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¬(hβ0)β‘γsinβ‘γ(( π₯ + β ) + 1 ) βγ sinγβ‘γ (π₯ + 1)γ γ/hγ = limβ¬(hβ0)β‘γ(π ππ( π₯ + 1 + β) βγ sinγβ‘γ ( π₯ + 1 )γ)/hγ Using sin A β sin B = 2 cos ((π΄ + π΅)/2) sin ((π΄ β π΅)/2) = limβ¬(hβ0)β‘γ(2 cosβ‘(( ( π₯ + 1 + β ) + ( π₯ + 1))/2).sinβ‘((( π₯ + 1 + β ) β ( π₯ + 1 ))/2))/hγ = limβ¬(hβ0)β‘γ(2 cosβ‘γ(( 2(π₯ + 1 )+ β))/2γ . γsin γβ‘γβ/2γ)/hγ = limβ¬(hβ0)β‘γ(cosβ‘γ(( 2(π₯ + 1 )+ β))/2γ . γsin γβ‘γβ/2γ)/(β/2)γ = limβ¬(hβ0)β‘γγcos γβ‘γ(( 2(π₯ + 1 )+ β))/2γ . sinβ‘γ β/2γ/(( β)/2)γ = limβ¬(hβ0)β‘γcos γβ‘γ(( 2(π₯ + 1 )+ β))/2γ .(π₯π’π¦)β¬(π‘βπ) πππβ‘γ π/πγ/(( π)/π) = limβ¬(hβ0)β‘γcos γβ‘γ(( 2(π₯ + 1 ) + β))/2γ .π = limβ¬(hβ0)β‘γcos γβ‘γ(( 2(π₯ + 1 )+ β))/2γ Putting h = 0 =γ cos γβ‘γ(( 2(π₯ + 1 ) + 0))/2γ = cos ( x + 1) Hence fβ (x) = cos (x + 1) Using (πππ)β¬(π₯β0)β‘γ π ππβ‘π₯/π₯γ=1 Replacing x by β/2 β (πππ)β¬(ββ0) π ππβ‘γ β/2γ/(( β)/2) = 1
Miscellaneous (Term 1 and Term 2)
Misc 1 (ii) Important
Misc 1 (iii) You are here
Misc 1 (iv) Important
Misc 2
Misc 3 Important
Misc 4 Important
Misc 5
Misc 6 Important
Misc 7
Misc 8 Important
Misc 9 Important
Misc 10
Misc 11
Misc 12 Important
Misc 13
Misc 14 Important
Misc 15
Misc 16
Misc 17 Important
Misc 18 Important
Misc 19
Misc 20 Important
Misc 21
Misc 22 Important
Misc 23
Misc 24 Important
Misc 25
Misc 26
Misc 27 Important
Misc 28 Important
Misc 29 Important
Misc 30 Important
Miscellaneous (Term 1 and Term 2)
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