Inverse trigonometric function

noypihenyo
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Find dy
dx


1.) y = In __x2 (x+1)___
(x + 2)3

2.) y = x3 ( 3lnx-1)

3.) y = __cos6 2x__
(1-sin2x)3


4.) y = __tan 2x__
1- cot 2x


5.) y = x e (exponent pa po ng e) sin2x


6.) Arc tan ____x_______
a- √ a2-x2


7.) __x_______ _ Arc sin _x_
a-√ a2-x2 a

_________
8.) Arc tan /__3x - 4_
√ 4
_______
9.) Arc cos ( 1 - _x_+ √ 2ax-x2
a

pls help i need the answer October 8
 
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3) and 4) are not about inverse functions. Show us what you have tried that didn't work.

For the rest, there is a clever trick you can apply for all of them. Say I want to find the derivative of y(x) = ln(x). since ln(x) is the inverse function of e^x=exp(x), I begin by "taking both sides into the argument of exp(x)", such that the equation becomes e^y=e^{ln(x)}=x. And now I differentiate both side wrt x: \frac{d}{dx}e^y=\frac{d}{dx}x. I use the chain rule on the left hand side: e^y\frac{dy}{dx} = 1, or, now replacing y by ln(x): \frac{dy}{dx} = 1/e^{ln(x)}=1/x. I suggest you study this exemple until you understand each step. Then, try one of your problems. If you get stuck, you can ask us for help by writing exactly what you've done and where you're stuck.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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