Deriving f(x)=sec(x^2)

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I need help finding the derivative of f(x)=secx^2 using the double angle formula. I know that the answer is 2xsec(x^2)tan(x^2) but I don't understand how they got the answer. 

Aug 24th, 2015

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Notice how a "co-(something)" trig ratio is always the reciprocal of some "non-co" ratio. You can use this fact to help you keep straight that cosecant goes with sine and secant goes with cosine.

sin2(t) + cos2(t) = 1  tan2(t) + 1 = sec2(t)  1 + cot2(t) = csc2(t)

The above, because they involve squaring and the number 1, are the "Pythagorean" identities. You can see this clearly if you consider the unit circle, where sin(t) = y, cos(t) = x, and the hypotenuse is 1.

sin(–t) = –sin(t)  cos(–t) = cos(t)  tan(–t) = –tan(t)

Notice in particular that sine and tangent are odd function, while cosine is an even function

angle sum and -Difference Identities

sin(α + β) = sin(α)cos(β) + cos(α)sin(β)
sin(α – β) = sin(α)cos(β) – cos(α)sin(β)
cos(α + β) = cos(α)cos(β) – sin(α)sin(β)
cos(α – β) = cos(α)cos(β) + sin(α)sin(β)
double angle Identities

sin(2x) = 2sin(x)cos(x)

cos(2x) = cos2(x) – sin2(x) = 1 – 2sin2(x) = 2cos2(x) – 1


Please let me know if you need any clarification. I'm always happy to answer your questions.
Aug 24th, 2015

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