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Differentiation of Trigonometric Functions : arctan, arcsec, arccos, arccot and arccsc

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

Solution Preview

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 ...

Solution Summary

Differentiation of Trigonometric Functions (arctan, arcsec, arccos, arccot and arccsc) is investigated.

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