Differentiate f(x)=tan(x^2+1).

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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You are only allowed to ask one question at a time. I am responding to your first question.

f(x) = tan(x^2 + 1)

We need to determine f'(x)

f'(x) = [sec (x^2 + 1)]^2 * (2x)

The required derivative of tan(x^2 + 1) = [sec (x^2 + 1)]^2 * (2x)

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll write tan(x^2 + 1) = sin(x^2 + 1)/cos (x^2 + 1)

We'll apply quotient rule and chain rule:

f'(x) = [sin(x^2 + 1)]'*cos (x^2 + 1) - sin(x^2 + 1)*[cos (x^2 + 1)]'/[cos (x^2 + 1)]^2

f'(x) = 2x*cos (x^2 + 1)*cos (x^2 + 1) + 2xsin(x^2 + 1)*sin(x^2 + 1)/[cos (x^2 + 1)]^2

f'(x) = 2x{[cos (x^2 + 1)]^2 + [sin (x^2 + 1)]^2}/[cos (x^2 + 1)]^2

But, from Pythagorean identity, [cos (x^2 + 1)]^2 + [sin (x^2 + 1)]^2 = 1

f'(x) = 2x/[cos (x^2 + 1)]^2

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