- #1
lamerali
- 62
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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!
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.
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.