Derivatives of trigonometric functions

[lamerali]

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.

 

[Dick]

You started going wrong in 2. Take a hard look at some examples of using the chain rule. Here for example it would say (f(x)^3)'=3*f(x)^2*f'(x). Apply this to the case where f(x)=cos(pi*x).

 

[lamerali]

so would question 2 be...

4 cos[tex]^{2}[/tex] ([tex]\pi[/tex] x)

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

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

 

[Dick]

That's it.

 

[lamerali]

great thank you!

as for question 3
y = 2x( [tex]\sqrt{x}[/tex] - cotx)

my second attempt at an answer:
y1 = 2( [tex]\sqrt{x}[/tex] - cotx) + 2x([tex]\frac{1}{2}[/tex] x[tex]^{- \frac{1}{2}}[/tex] + csc[tex]^{2}[/tex] x
= 2( [tex]\sqrt{x}[/tex] - cotx) + [tex]\sqrt{x}[/tex] + 2x csc[tex]^{2}[/tex] x


is this anywhere near correct?
thanks

i am not sure how to get on with question 4: y = tan [tex]^{2}[/tex] (cosx)

thank you for the help!

 

[Dick]

That's also correct. You could combine the square roots by expanding out, but that doesn't make it incorrect. For 4, it might help to rewrite it a little. Let's define sqr(x)=x^2, ok? Then that is sqr(tan(cos(x))). If you use the chain rule twice you can show that (f(g(h(x)))'=f'(g(h(x))*g'(h(x))*h'(x). Do you see how that's working? You keep taking the derivative of the outside function evaluated at the inside function times the derivative of the inside function.

 

[lamerali]

alright...i'm not sure i got this one but here it goes:

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

y1 = 2tan cosx + 2 sec[tex]^{2}[/tex]cosx - sin tan[tex]^{2}[/tex]

how does it look? :|

 

[lamerali]

also are question five and six okay? i don't see how i can come up with any other solutions.

 

[Dick]

alright...i'm not sure i got this one but here it goes:

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

y1 = 2tan cosx + 2 sec[tex]^{2}[/tex]cosx - sin tan[tex]^{2}[/tex]

how does it look? :|

It looks kind of incoherent. You can either use full tex stuff or you can do what I usually do and try to approximate it with lots of characters and parenthesis. But I don't understand that at all.

 

[Dick]

also are question five and six okay? i don't see how i can come up with any other solutions.

I think 6 is ok. 5 is awful. (1/f(x))' is not equal to 1/(f'(x)), is it?

 

[lamerali]

okey i believe i figured question 5 out:

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

y1 = (1 + tanx)[tex]^{-1}[/tex]
= (-1)(1 + tanx)(sec[tex]^{-2}[/tex] sec [tex]^{2}[/tex] x
= - [tex]\frac{(sec^2)x}{(1 + tanx)^2}[/tex]
I am still unsure where i am going with question 4 but here is my zillionth attempt :D

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

my answer:

y1 = 2tan(cosx)sec[tex]^{2}[/tex]x(cosx)(-sinx)
y1 = -2sin x tan(cosx) sec[tex]^{2}[/tex](cosx)


thank you. i appreciate all the help! :D

 

[Dick]

They both look correct. You are improving a lot!

 

[lamerali]

Thank you sooo much! couldn't have done that without you! THANKS for alllll the help! :D