Expert: Paul Klarreich - 7/31/2009

Question
A curve has equation(x+1)^2 (y+2)sin x+(x+1)(y+2)^2cos x=4.Show that the curve passes through (0,0) and determine the value of dy/dx at (0,0). I was thinking product rule but I dont know which values to assign the u&v and which ones to leave out.Thanks in advance!

Answer
Questioner: Lucilla Joyce
Country: Norway
Category: Calculus
Private: No
Subject: Differentiation
Question: A curve has equation(x+1)^2 (y+2)sin x+(x+1)(y+2)^2cos x=4.Show that the curve passes through (0,0) and determine the value of dy/dx at (0,0). I was thinking product rule but I dont know which values to assign the u&v and which ones to leave out.Thanks in advance!
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(x + 1)^2 (y+2)sin x + (x + 1)(y + 2)^2 cos x = 4

Show that the curve passes through (0,0)

>> This is routine substitution.

and determine the value of dy/dx at (0,0)

>> This is not -- implicit differentiation, THEN routine substitution.

But you have terms that involve products, so you will need the product rule, A LOT!  You should do each term separately, then put them together.
...........................
First term:

(x + 1)^2 (y+2)sin x

u = (x + 1)^2,  v = (y + 2)sin x, and the "v" is itself a product.

u' = 2(x + 1),

v' = (y + 2) cos x + dy/dx sin x

Derivative1 = (x + 1)^2 [(y + 2) cos x + dy/dx sin x] + (y + 2)sin x[2(x + 1)]
..........................
Second term.(BTW, D(right side) = 0)
(x + 1)(y + 2)^2 cos x

Now let's do  u = (x + 1) cos x,  v = (y + 2)^2

u' = -(x + 1)sin x + cos x

v' = 2(y + 2) dy/dx

Derivative2 = (x + 1) cos x (2(y + 2) dy/dx) + (y + 2)^2(-(x + 1)sin x + cos x)
.............................
Now your left side is Derivative1 + Derivative2.  [What?  Someone told you this was going to be easy?]

Derivative1 = (x + 1)^2 [(y + 2) cos x + dy/dx sin x] + (y + 2)sin x[2(x + 1)]
Derivative2 = (x + 1) cos x (2(y + 2) dy/dx) + (y + 2)^2(-(x + 1)sin x + cos x)

AT (0,0):

Derivative1 = 2  
Derivative2 =  (2(2) dy/dx) + (2)^2

2 + 4 y' + 4 = 0

4y' = -6

y' = -3/2
........................................

Another way to do this is to rewrite your equation:

(x + 1)^2 (y+2)sin x + (x + 1)(y + 2)^2 cos x = 4

into:
                                   4
(x + 1) sin x + (y + 2) cos x = ----------
                               (x+1)(y+2)

Now the left side is much simpler.  The RHS will involve the quotient and product rules, but that won't be so bad.
        - 4[(x+1)dy/dx + y + 2]
Deriv3 = -----------------------
              (x+1)^2(y+2)^2

Deriv1 = (x + 1)cos x + sin x

Deriv2 = -(y + 2)sin x + cos x dy/dx

So you would have, at (0,0):

        - 4[dy/dx + 2]
Deriv3 = -------------- = - dy/dx - 2
              4

Deriv1 = 1

Deriv2 = dy/dx

1 + dy/dx = - dy/dx - 2

2 dy/dx = -3,  etc.