1.Find an equation of the line tangent to sin(xy)=y at the point (pi/2,1) 2.Find

Question

1.Find an equation of the line tangent to sin(xy)=y at the point (pi/2,1)

2.Find DxY for the function x^2+2x^2y+3xy=7 by implicit differentation

3. Use linerization of y=sqr tx to

a)approximate sqrt 402

b)compare with your calculators answer for sqrt 402

c) to how many decimals places do they agree

Explanation / Answer

1) sin(xy)=y differentiate it we get cos(xy) *[ y + x dy/dx] = dy/dx ==> dy/dx = -[y cos(xy)]/[x cos(xy) -1] slope at (pi/2,1) = dy/dx = -[y cos(xy)]/[x cos(xy) -1] = -[1* cos(pi/2)]/[pi/2 * cos(pi/2) -1] = 0 so, eq. of the tangent is y = 1 2) x^2+2x^2y+3xy=7 differentiate 2x + 4x y + 2x^2 dy/dx + 3y +3x dy/dx = 0 ==> dy/dx = [-2x -4xy -3y]/[2x^2 +3x]

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