Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

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Ex 5.2, 5 Differentiate the functions with respect to π₯ : sinβ‘γ (ππ₯ + π)γ/cosβ‘γ (ππ₯ + π)γ Let π¦ = sinβ‘γ (ππ₯ + π)γ/cosβ‘γ (ππ₯ + π)γ Let π’ = sinβ‘γ (ππ₯+π)γ & π£=cosβ‘γ (ππ₯+π)γ β΄ π = π/π We need to find derivative of π¦ π€.π.π‘.π₯ ππ¦/ππ₯ = (π’/π£)^β² ππ¦/ππ₯ = (π’^β² π£ β γ π£γ^β² π’)/π£^2 Finding πβ π’=sinβ‘γ (ππ₯+π)γ Derivative of π’ π€.π.π‘.π₯ ππ’/ππ₯ =π(sinβ‘γ (ππ₯+π)γ )/ππ₯ γ=cos γβ‘(ππ₯+π) . π(ππ₯ + π)/ππ₯ γ=cos γβ‘(ππ₯+π) (π +0) γ=π πππ γβ‘(ππ+π) Finding πβ π£=cosβ‘γ (ππ₯+π)γ Derivative of π£ π€.π.π‘.π₯ ππ£/ππ₯ = (π(cosβ‘γ (ππ₯ + π)γ )^β² )/ππ₯ γ=βsiπ γβ‘(ππ₯+π) . π(ππ₯ + π)/ππ₯ γ=βsin γβ‘(ππ₯+π) (π+0) γ=βπ πππ γβ‘(ππ+π) Now, ππ/ππ = (π^β² π β γ πγ^β² π)/π^π = (γa cos γβ‘γ(ππ₯ + π) .γ cosγβ‘γ (ππ₯ + π)γ β (βπ γsin γβ‘γ(ππ₯ + π) γ ) γ (γsin γβ‘γ(ππ₯ + π)γ ) )/(cosβ‘γ (ππ₯ + π)γ )^2 = (γa cos γβ‘γ(ππ₯ + π) .γ cosγβ‘(ππ₯ + π) + π . γsin γβ‘(ππ₯ + π) γ sinβ‘γ(ππ₯ + π)γ )/cos^2β‘(ππ₯ + π) = (γa cos γβ‘γ(ππ₯ + π) .γ cosγβ‘γ (ππ₯ + π)γ γ )/cos^2β‘(ππ₯ + π) + (π . γsin γβ‘γ(ππ₯ + π) γ. sinβ‘γ (ππ₯ + π)γ )/cos^2β‘(ππ₯ + π) = a cos (ππ₯+ π) π/πππβ‘γ (ππ + π)γ + π . γsin γβ‘(ππ₯+π) . (γπππ γβ‘γ(ππ + π)γ )/πππβ‘γ (ππ + π)γ π/πππβ‘γ (ππ + π)γ = π πππ (ππ₯+ π) .πππβ‘γ(ππ₯+π)γ + π . γπππ γβ‘(ππ₯+π).γπππ γβ‘(ππ₯+π).πππβ‘(ππ₯+π)