Differentiation of Trigonometric Functions : arctan, arcsec, arccos, arccot and arccsc
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Please verify each differentiation formula and explain how.
(a) d/dx[arctan u] = u'/1 + u^2
(b) d/dx[arcsec u] = u'/lul (square root of u^2 -1)
(c) d/dx[arccos u] = -u'/square root of 1 - u^2
(d) d/dx[arccot u] = -u'/1- u^2
(e) d/dx[arccsc u] = -u'/lul (square root of u^2 - 1
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Solution Summary
Differentiation of Trigonometric Functions (arctan, arcsec, arccos, arccot and arccsc) is investigated.
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Hi , Here is the your questions
a) d/dx[arctan u] = u'/1 + u^2
(b) d/dx[arcsec u] = u'/lul (square root of u^2 -1)
(c) d/dx[arcos u] = -u'/square root of 1 - u^2
(d) d/dx[arccot u] = -u'/1- u^2
(e) d/dx[arccsc u] = -u'/lul (square root of u^2 - 1
(a) Here you want to prove or verify how d/dx(arc tanu) = u'/1+u^2
Arc tan u is nothing but Tan^-1 (u)
Take y= tan^-1(u)
Then tan y = u
Differentiating both sides with respect to x, we get
sec^2(y). dy/dx = du/dx
du/dx
dy/dx = _______
sec^2( y)
We know the identity 1+ tan^2 (y) = sec^2 (y)
du/dx
dy/dx = _______
1+ tan^2( y)
We took y= tan^-1 (u) and u = tan y
so d/dx ( tan^-1 (u)) = u'/ 1+ u^2
Hence verified.
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(b)d/dx[arcsec u] = u'/lul ...
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