Math
posted by Ifi .
dx/dy cos x sin y + x^5 = 11
i got dx/dy= (cos x cos y)/ (sin x sin y +5x^4)
But the wolfram got
(1 cos x cos y)/(5x^4sin y sin x), where is the 1 come from?

Math 
Reiny
so we are differentiating
cosx siny + x^5 = 11 implicitly ?
cosx(cosy) dy/dx + siny(sinx) + 5x^4 = 0
dy/dx = (sinx siny  5x^3)/(cosx cosy)
which agrees with my Wolfram result
http://www.wolframalpha.com/input/?i=find+dy%2Fdx+for+%28cos%28x%29%29%28sin%28y%29%29+%2B+x%5E5+%3D+11 
Math 
Reiny
just noticed you wanted dx/dy (strange)
cosx siny + x^5 = 11
cosx(cosy) + siny(sinx) dx/dy = 5x^4 dx/dy = 0
dx/dy (5x4  sinxsiny) =  cosxcosy
dx/dy = cosxcosy/(5x^4  sinxsiny)
the same as yours. 
Math 
Ifi
At reiny .... Thanks
Respond to this Question
Similar Questions

trig
Reduce the following to the sine or cosine of one angle: (i) sin145*cos75  cos145*sin75 (ii) cos35*cos15  sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b)  sin(a)sin)(b) (1)The quantity … 
tigonometry
expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by b and using that cos(b)= cos(b) sin(b)= sin(b) gives: sin(ab) = sin(a)cos(b)  cos(a)sin(b) … 
Trig
Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v  u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos … 
algebra
Can someone please help me do this problem? 
precal
Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x 48 
Mathematics  Trigonometric Identities
Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y … 
TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1  (3/4)sin^2 2x work on one side only! Responses Trig please help!  Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 … 
Precal
I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1  sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 =  … 
Trig
Find sin(s+t) and (st) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(1/5)Sin(3/5) = 0.389418 Sin(st) =sin(s)cos(t)  cos(s)sin(t) =sin(3/5)cos(1/5)  cos(1/5)sin(3/5) … 
Math
Show that for real x that {[cos x + 2 sin x + 1]/[cos x + sin x] } cannot have a value between 1 and 2. Let y = [(cos x+2 sin x + 1)/(cos x + sin x) ] y(cos x + sin x) = (cos x + 2 sin x + 1) sin x(y2) + cos x(y1)=1 , I just feel …