find the eqn of normal to the curve
y=(1+x)^y +sin inverse (sin^2 x) at x=0

Question

find the eqn of normal to the curve 
y=(1+x)^y +sin inverse (sin^2 x) at x=0

aravindg · Answer

@apoorvk , @Mani_Jha

experimentx · Answer

looks like a nose http://www.wolframalpha.com/input/?i=graph+y%3D%281%2Bx%29%5Ey+%2Bsin+inverse+%28sin%5E2+x%29

aravindg · Answer

@asnaseer , @satellite73 , @TuringTest

mani_jha · Answer

Find dy/dx at x=0
Let the point be (0,b)
Slope of normal:
$$-1/(dy/dx)$$
Equation of normal:
$$(y-b)=(-1/(dy/dx))(x-0)$$
Got it?

aravindg · Answer

hmm. w8

experimentx · Answer

this doesn't look easy.

asnaseer · Answer

Mani_jha is correct - spot on!

aravindg · Answer

i will try tht and let u knw if i get stuck..anyway thx all

experimentx · Answer

I think the question is more like, solve for y, then find dy/dx 

put the values of x,y on dy/dx then calculate the normal.

experimentx · Answer

looks like it was that point mani mentioned was 0,1
I got crazy value for dy/dx  ... please check it out.

asnaseer · Answer

experimentX: you can just use implicit derivative - no need to solve for y first.

experimentx · Answer

still, the dy/dx will be interms of y, so y is needed at that point,

asnaseer · Answer

dy/dx will most likely be near 0 at x=0 by the looks of the curve

experimentx · Answer

yeah ... but i got 1, y=1, ... looks like there must some error.

experimentx · Answer

*m=1

asnaseer · Answer

$$y=(1+x)^y+\sin^{-1}(\sin^2(x))$$$$y'=y(1+x)^{y-1}*y'*1+\frac{\sin(2x)}{\sqrt{1-\sin^4(x)}}$$which leads to:$$y'=\frac{\sin(2x)}{(1+y(1+x)^{y-1})\sqrt{1-\sin^4(x)}}$$

asnaseer · Answer

unless I have made a mistake somewhere, this gives y'=0 at x=0

experimentx · Answer

yeah ... it's zero, looks like i made mistake in taking dy/dx

experimentx · Answer

does (x+1)^y work with chain rule??

asnaseer · Answer

I believe so

aravindg · Answer

ready for the next qn?

experimentx · Answer

I thought it was some what like 

x^x, or (1+x)^f(x)  <--- though I am not quite sure on this.

asnaseer · Answer

just checked with the wolf - you could be right experimentX: http://www.wolframalpha.com/input/?i=differentiate+y%3D%281%2Bx%29^y+with+respect+to+x

experimentx · Answer

i hope so.. still not getting the value of m. let me revise.

asnaseer · Answer

the derivative at x=0 works out to be equal to y.

experimentx · Answer

?? I got the same result. dy/dx = 1 + 0

experimentx · Answer

might be due to this crazy function http://www.wolframalpha.com/input/?i=graph+y+%3D+%281%2Bx%29%5Ey