# Math Help - Trig derivatives

1. ## Trig derivatives

ok so I think this is using chain rule again but i'm stuck at simplifying

y= ((1+cosx)/(sinx))^(-1)

i'm at

-1 (1+cosx/sinx) ^(-2) times (-sinx^2 - cos x +cos x^2)/sinx^2

and also

y=sec^(2) pi x
ok so there isn't any brackets around pix its just right beside sec^2 so i dont know how the chain rule would work?

2. Originally Posted by rmn
ok so I think this is using chain rule again but i'm stuck at simplifying

y= ((1+cosx)/(sinx))^(-1)

i'm at

-1 (1+cosx/sinx) ^(-2) times (-sinx^2 - cos x +cos x^2)/sinx^2

and also

y=sec^(2) pi x
ok so there isn't any brackets around pix its just right beside sec^2 so i dont know how the chain rule would work?
Ummm....
$y = \left ( \frac{1 + cos(x)}{sin(x)} \right ) ^{-1} = \frac{sin(x)}{1 + cos(x)}$

So
$\frac{dy}{dx} = \frac{cos(x) \cdot (1 + cos(x)) - sin(x) \cdot -sin(x)}{(1 + cos(x))^2}$

No need to do the chain rule, just use the quotient rule.

-Dan

3. thanks
for
y=sec^(2) pi x
do i just use the multiplication rule?

4. Originally Posted by rmn
thanks
for
y=sec^(2) pi x
do i just use the multiplication rule?
no, the chain rule (note, you would have to do it twice, once for sec^2 and one for pi x, post your solution so i can see if you get what i'm saying)

5. Originally Posted by Jhevon
no, the chain rule (note, you would have to do it twice, once for sec^2 and one for pi x, post your solution so i can see if you get what i'm saying)
theres no brackets around the pix though?

well heres what i did

sec^2 tan^2 (pi x) (pi)

6. Originally Posted by rmn
theres no brackets around the pix though?
what do you mean? (pi x) is what the secant is operating on, it is a linear function of x.

sec^2 tan^2 (pi x) (pi)
that is incorrect. recall the chain rule: $\frac d{dx}f(g(x)) = f'(g(x))g'(x)$

7. Originally Posted by Jhevon
what do you mean? (pi x) is what the secant is operating on, it is a linear function of x.

that is incorrect. recall the chain rule: $\frac d{dx}f(g(x)) = f'(g(x))g'(x)$

is pi x g(x)?

8. Originally Posted by rmn
is pi x g(x)?
no, $\sec \left( \pi x \right)$ is $g(x)$, the thing is, g(x) is itself a composite function, which is why you need the chain rule twice

9. Originally Posted by Jhevon
no, $\sec \left( \pi x \right)$ is $g(x)$, the thing is, g(x) is itself a composite function, which is why you need the chain rule twice

2 sec pi x sec ^2 tan^2 pi x ?

10. Originally Posted by rmn
2 sec pi x sec ^2 tan^2 pi x ?
what is the derivative of sec(x)?

11. Originally Posted by Jhevon
what is the derivative of sec(x)?
secxtanx

12. Originally Posted by rmn
secxtanx
correct, so where did sec^2 tan^2 (pi x) come from?

13. $\frac d{dx} \sec^2 ( \pi x) = 2 \pi \sec (\pi x) \cdot \sec (\pi x) \tan (\pi x) = 2 \pi \sec^2 (\pi x) \tan (\pi x)$

now, can you tell me why?

14. Originally Posted by Jhevon
$\frac d{dx} \sec^2 ( \pi x) = 2 \pi \sec (\pi x) \cdot \sec (\pi x) \tan (\pi x) = 2 \pi \sec^2 (\pi x) \tan (\pi x)$

now, can you tell me why?

in the question there is no brackets around pi x but u put brackets around it, or does it matter if there is

15. Originally Posted by rmn
in the question there is no brackets around pi x but u put brackets around it, or does it matter if there is
i put it for clarification, so you know what the secant is operating on

Page 1 of 2 12 Last