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Derivatives of sinusoidal functions help?

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y=secx/ cos^2x I tried it but my answers is different than the back? Help

asked Mar 30, 2013 in CALCULUS by andrew Scholar

1 Answer

+1 vote

y = secx / cos2x

Recall : Trigonometry formulas secx = 1 / cosx then 1 / cos2x = sec2x

There fore y = secx (sec2x)

Simplify(power rule)

y = sec3x

Diferenciate with respective x to each side

Recall : Derivative formulas d/dx(xn) = nxn-1 and d/dx(secx) = secxtanx

dy /dx = 3sec2x secxtanx .1

dy / dx = 3sec3x sinx / cosx      (tanx = sinx / cosx)

dy / dx = 3sec3x sinx secx

dy / dx = 3sec4x sinx

dy / dx = 3sinx / cos4x

dy / dx  =3sinx / (cos2x)2       (cos2x = 1 - sin2x)

dy / dx = 3sinx / (1 - sin2x)2

dy / dx =3sinx / (1 - 2sin2x + sin4x) .

answered Mar 30, 2013 by diane Scholar

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