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
= γπ¬ππ γβ‘γ(πππ§β‘βπ)γ γπππ§ γβ‘γ(πππβ‘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₯)
= (πππβ‘γ(πππβ‘βπ γ)πππβ‘γ(πππβ‘βπ γ)γπππγ^πβ‘βπ )/(πβπ)

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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