Hi, i'm working with finding the derivatives of trigonometric functions but i'm not confidant with some of my answers. if someone would go over these derivatives i would appreciate it. thanks in advance!(adsbygoogle = window.adsbygoogle || []).push({});

determine [tex]\frac{dy}{dx}[/tex] . do not simplify.

question 1

y = sec [tex]\sqrt[3]{x}[/tex]

my answer:

y1 = sec [tex]\sqrt[3]{x}[/tex] tan[tex]\sqrt[3]{x}[/tex]

= sec [tex]\sqrt[3]{x}[/tex] tan[tex]\sqrt[3]{x}[/tex] [tex]\frac{1}{3}[/tex] x[tex]^{- \frac{2}{3}}[/tex]

= [tex]\frac{sec \sqrt[3]{x} tan \sqrt[3]{x}}{ 3\sqrt[3]{x^{2}}}[/tex]

question 2

y = 4cos[tex]^{3}[/tex] ([tex]\pi[/tex] x)

my answer:

y1 = 12(-sin [tex]^{2}[/tex] [tex]\pi[/tex] x) [tex]\pi[/tex]

= -12([tex]\pi[/tex] sin [tex]^{2}[/tex] [tex]\pi[/tex] x

question 3

y = 2x([tex]\sqrt{x}[/tex] - cot x)

my answer:

y1 = 2([tex]\sqrt{x}[/tex] - cot x) + (2x) [tex]\frac{1}{2}[/tex] x[tex]^{- \frac{1}{2}}[/tex] - (-csc [tex]^{2}[/tex] x))

= 2([tex]\sqrt{x}[/tex] - cot x) + (2x) [tex]\frac{csc^{2}x}{2\sqrt{x}}[/tex]

= 2([tex]\sqrt{x}[/tex] - cot x) + [tex]\frac{x csc^{2}x}{\sqrt{x}}[/tex]

question 4

y = tan [tex]^{2}[/tex] (cos x)

my answer

y1 = 2sec[tex]^{2}[/tex] (-sinx)

question 5

y = [tex]\frac{1}{1 + tanx}[/tex]

my answer

y1 = [tex]\frac{1}{sec^{2}x}[/tex]

question 6

sinx + siny = 1

cos x + cos y [tex]\frac{dy}{dx}[/tex] = 0

[tex]\frac{dy}{dx}[/tex] = - [tex]\frac{cosx}{cosy}[/tex]

I'm not sure how i did with these. if someone could overlook them i'd be very greatful.

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# Homework Help: Derivatives of trigonometric functions

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