Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12



Last updated at Jan. 3, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12
Transcript
Ex 5.2, 4 (Introduction) Differentiate the functions with respect to x sec (tan ( βπ₯ )) Finding derivative of sec x Let π¦=secβ‘π₯ π¦ = 1/cosβ‘π₯ Let π’ = 1 & π£ = cosβ‘π₯ So, y = π’/π£ (y)β = (π’/π£)^β² π¦β² = (π’^β² π£ β π£^β² π’)/π£^2 Putting values π¦β² = (γ(1)γ^β² cosβ‘π₯ β γ(cosβ‘γπ₯ γ)γ^β² 1)/cos^2β‘π₯ yβ = (0 γ(cosγβ‘γπ₯)γ β (γβsinγβ‘γπ₯) 1γ)/(γπππ γ^2 π₯) = 0 + sinβ‘π₯/(γπππ γ^2 π₯) = sinβ‘π₯/cosβ‘π₯ . 1/cosβ‘π₯ = tanβ‘π₯.secβ‘π₯ Hence, (πππβ‘π )β =πππβ‘π πππβ‘π Ex 5.2, 4 Differentiate the functions with respect to x sec (tan ( βπ₯ )) Let π¦ = sec (tan βπ₯ ) We need to find Derivative of π¦ i.e. π¦β = (secβ‘γγ(tanγβ‘βπ₯)γ )^β² = γsec γβ‘γ(tanβ‘βπ₯)γ γtan γβ‘γ(tanβ‘βπ₯)γ (tanβ‘βπ₯ )^β² = γsec γβ‘γ(tanβ‘βπ₯)γ γtan γβ‘γ(tanβ‘βπ₯)γ. ("sec2 " βπ₯ " . " (βπ₯)^β²) = γsec γβ‘γ(tanβ‘βπ₯)γ γtan γβ‘γ(tanβ‘βπ₯)γ. sec2 " " βπ₯ Γ 1/(2βπ₯) = (secβ‘γ(tanβ‘βπ₯ γ)secβ‘γ(tanβ‘βπ₯ γ)sec^2β‘βπ₯ )/(2βπ₯) Hence π π/π π = (πππβ‘γ(πππβ‘βπ γ)πππβ‘γ(πππβ‘βπ γ)γπππγ^πβ‘βπ )/(πβπ)
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