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 $$\frac{dy}{dx}$$ . do not simplify.

question 1
y = sec $$\sqrt[3]{x}$$

my answer:
y1 = sec $$\sqrt[3]{x}$$ tan$$\sqrt[3]{x}$$
= sec $$\sqrt[3]{x}$$ tan$$\sqrt[3]{x}$$ $$\frac{1}{3}$$ x$$^{- \frac{2}{3}}$$

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

question 2

y = 4cos$$^{3}$$ ($$\pi$$ x)

my answer:

y1 = 12(-sin $$^{2}$$ $$\pi$$ x) $$\pi$$

= -12($$\pi$$ sin $$^{2}$$ $$\pi$$ x

question 3

y = 2x($$\sqrt{x}$$ - cot x)

my answer:
y1 = 2($$\sqrt{x}$$ - cot x) + (2x) $$\frac{1}{2}$$ x$$^{- \frac{1}{2}}$$ - (-csc $$^{2}$$ x))

= 2($$\sqrt{x}$$ - cot x) + (2x) $$\frac{csc^{2}x}{2\sqrt{x}}$$
= 2($$\sqrt{x}$$ - cot x) + $$\frac{x csc^{2}x}{\sqrt{x}}$$

question 4

y = tan $$^{2}$$ (cos x)

my answer
y1 = 2sec$$^{2}$$ (-sinx)

question 5
y = $$\frac{1}{1 + tanx}$$

my answer
y1 = $$\frac{1}{sec^{2}x}$$

question 6
sinx + siny = 1
cos x + cos y $$\frac{dy}{dx}$$ = 0

$$\frac{dy}{dx}$$ = - $$\frac{cosx}{cosy}$$


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$$^{2}$$ ($$\pi$$ x)

12 cos $$^{2}$$ ($$\pi$$ x) (-sin ($$\pi$$ x)) ($$\pi$$)

(-sin ($$\pi$$ x)) 12 $$\pi$$ cos $$^{2}$$ ($$\pi$$ x)

 

[Dick]

That's it.

 

[lamerali]

great thank you!

as for question 3
y = 2x( $$\sqrt{x}$$ - cotx)

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


is this anywhere near correct?
thanks

i am not sure how to get on with question 4: y = tan $$^{2}$$ (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 $$^{2}$$ (cos x)

y1 = 2tan cosx + 2 sec$$^{2}$$cosx - sin tan$$^{2}$$

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 $$^{2}$$ (cos x)

y1 = 2tan cosx + 2 sec$$^{2}$$cosx - sin tan$$^{2}$$

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 = $$\frac{1}{1+tanx}$$

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

y = tan$$^{2}$$(cos x)

my answer:

y1 = 2tan(cosx)sec$$^{2}$$x(cosx)(-sinx)
y1 = -2sin x tan(cosx) sec$$^{2}$$(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