# Thread: derivatives of trigonometric functions

1. ## derivatives of trigonometric functions

find dy/dx of

y= (cosx/x) + (x/cosx)

i got up to

((xsinx-cosx)/x^2 ) + ((cosx+sinx)/cosx^2))

and i don't know how to simplify it.

2. Originally Posted by rmn
find dy/dx of

y= (cosx/x) + (x/cosx)

i got up to

((xsinx-cosx)/x^2 ) + ((cosx+sinx)/cosx^2))

and i don't know how to simplify it.
perhaps it would be easier to simplify if you had combined the fractions first.

note that: $\displaystyle y = \frac {\cos^2 x + x^2}{x \cos x}$

Now, perform the quotient rule on that.

3. ok so I now have

((xcosx)(-sinx^2 + 2x) - (cosx^2+x^2)(x)(-sinx)+cosx) / (xcosx)^2

i'm confused of how to simplify.. like how do you multiply xcosx with stuff?

4. Originally Posted by rmn
ok so I now have

((xcosx)(-sinx^2 + 2x) - (cosx^2+x^2)(x)(-sinx)+cosx) / (xcosx)^2

i'm confused of how to simplify.. like how do you multiply xcosx with stuff?
that derivative is incorrect. you need the chain rule to derive things like $\displaystyle \cos^2 x$

this is my 42th post!!!!

5. after trying your way and my way, i found that the problem is generally easier to do your way. however, i do not see a worth-while way to simplify anything, so just leave the answer as:

$\displaystyle y' = \frac {-x \sin x - \cos x}{x^2} + \frac {\cos x + \sin x}{\cos^2 x}$

or if you feel like, you can combine these fractions to get: $\displaystyle \frac {-x \sin x \cos^2 x - \cos^3 x + x^2 \cos x + x^2 \sin x}{x^2 \cos^2 x}$

but i don't see any motivation for doing even that