Maths(Implicit functions)

2007-04-24 10:17 pm
find dy/dx,
arctan(x^2y) = xy^2

回答 (2)

2007-04-25 5:29 am
✔ 最佳答案
First differentiate both sides w.r.t x
LHS = 1/(1+(x^2y)^2) * d(x^2y)dx
= 1/(1+(x^2y)^2) * (2xy + x^2*dy/dx)

RHS = d(xy^2)/dx = y^2 + 2xy*dy/dx

Equating both sides,
1/(1+(x^2y)^2) * (2xy + x^2*dy/dx) = y^2 + 2xy*dy/dx
2xy/(1+(x^2y)^2) + [x^2/(1+(x^2y)^2)]*dy/dx) = y^2 + 2xy*dy/dx
2xy/(1+(x^2y)^2)-y^2 = [2xy-x^2/(1+(x^2y)^2)]*dy/dx
(2xy-y^2-x^4y^4)/(1+(x^2y)^2) = (2xy+2x^5y^3-x^2)/(1+(x^2y)^2)*dy/dx
dy/dx = (2xy-y^2-x^4y^4)/(2xy+2x^5y^3-x^2)
2007-04-25 5:50 pm
arctan(x²y) = xy²
x²y = tan(xy²)

L.H.S. = x²y
(d/dx)(x²y) = y(2x) + (x²)(dy/dx)

R.H.S.
tan(xy²) = sec²(xy²)(y² + x(2y(dy/dx)))
= y²sec²(xy²) + 2xysec²(xy²)(dy/dx)

therefore
y(2x) + (x²)(dy/dx) = y²sec²(xy²)+2xysec²(xy²)(dy/dx)
(x²)(dy/dx) - 2xysec²(xy²)(dy/dx) = y²sec²(xy²) - 2xy
(dy/dx)(x² - 2xysec²(xy²)) = y²sec²(xy²) - 2xy
(dy/dx) = (y²sec²(xy²) - 2xy) / (x² - 2xysec²(xy²))

after simplify
(dy/dx) = (y(y - 2xcos²xy²)) / (x(xcos²(xy²)-2y)) //


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