calc
posted by Chelsea .
1 + x = sin(xy^2)
find dy/dx by implicit differentiation
0 + 1 = cos(xy^2). (x)(2y)dy/dx + (y^2)(1)
1/((x)(2y)dy/dx) = cos(xy^2) + (y^2)
dy/dx = cos (xy^2) + (y^2)....
Can I just divide out the (x)(2y) and leave the dy/dx?

you needed brackets like this
0 + 1 = cos(xy^2)*( (x)(2y)dy/dx + (y^2)(1) )
since you need to get at the dy/dx and since it is inside the bracket, you must expand
1 = 2xy(cos(xy^2))dy/dx + y^2(cos(xy^2)
1  y^2(cos(xy^2) = 2xy(cos(xy^2))dy/dx
dy/dx = (1  y^2(cos(xy^2))/(2xy(cos(xy^2)))
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) … 
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 =  … 
Calc.
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x) (dy/dx)(dx/du)= … 
calculus
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x) (dy/dx)(dx/du)= … 
Calculus
Find dy/dx for y=sin(x+y) A. 0 B. (cos(x+y))/(1cos(x+y) C. cos(x+y) D. 1 E. None of these I know I'm supposed to use implicit differentiation but I'm not sure how to go about it with sin