# Math Help - Trig derivatives

1. 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?
Here is one way.
Sorry but I forgot already those chain rule, quotient rule, whatever rule. I just keep on differentiating until all the variables are differentiated, etc. I must be following those said rules although those rules are not in my mind any more.

y = [(1 +cosX) / sinX]^(-1)

dy/dX = -1[(1 +cosX) / sinX]^(-2) *({sinX[-sinX] -(1 +cosX)[cosX]} / sin^2(X))

dy/dX = (-1 / [(1 +cosX) / sinX]^2) * ({-sin^2(X) -cosX -cos^2(X)} / sin^2(X))

dy/dX = -(sin^2(X) / (1 +cosX)^2) * -({sin^2(X) +cos^2(X) +cosX} / sin^2(X))

The sin^2(X) cancell out; the sin^2(X) +cos^2(X) is converted to 1; the multiplication of negatives turns to positive,

dy/dX = (1 +cosX) / (1 +cosX)^2

dy/dX = 1 / (1 +cosX) ------------------------answer.

---------------------------------------------------------------

y = sec^2(pi*X)
Or,
y = [sec(pi*x)]^2

dy/dX = 2[sec(pi*X)] *[sec(pi*X)tan(pi*X)] *[pi]

dy/dX = (2pi)sec^2(pi*X)tan(pi*X) -----------------answer.

2. ## same

Iv also got a problem with derivative of trig functions, heres mine:

[sinInverse(x)/sqrt(1-x^2) dx] for range 0...0.5

am not sure how to solve that?

3. Originally Posted by taurus
Iv also got a problem with derivative of trig functions, heres mine:

[sinInverse(x)/sqrt(1-x^2) dx] for range 0...0.5

am not sure how to solve that?
are you sure you mean derivative? you have a dx attached to it. that means you should integrate

4. sorry yea i mean integrate?

5. $\int \frac {\arcsin x}{\sqrt{1 - x^2}}~dx$

note that $\frac d{dx} \arcsin x = \frac 1{\sqrt {1 - x^2}}$

thus, a simple substitution $u = \arcsin x$ will do the trick here.

try it out.

EDIT: whatever, i'll just do it

Let $u = \arcsin x$

$\Rightarrow du = \frac 1{\sqrt {1 - x^2}}~dx$

So our integral becomes

$\int u ~du = \frac 12u^2 + C = \frac 12 \arcsin^2 x + C$

6. just a quick question, where did the half come from?

7. Originally Posted by taurus
just a quick question, where did the half come from?
it is from the power rule for integrals. $\int x^n~dx = \frac {x^{n + 1}}{n + 1} + C$ for $n \ne -1$

thus $\int u~du = \frac {u^{1 + 1}}{1 + 1} + C = \frac 12u^2 + C$

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