Posted by Albert on Thursday, June 21, 2012 at 3:02pm.
normally you just have an equation like
y = 3x^2
and you take the derivative of both sides, to get
y' = 6x
Then you advance to using the chain rule and product rule, and you have something like
y = e^x*sin^2(6x)
and you take the derivative of both sides to get
y' = e^x(sin^2(6x)+12sin(6x)cos(6x))
implicit differentiation just uses these techniques. Take the derivative of both sides of the equation, using the chain rule, and remembering that dx/dx = x' = 1 and dy/dx = y'
So, for your problems,
a) the derivative of xy is
d/dx(xy) = x'y + xy' = y + xy'
d/dx(x^5y^2) = (5x^4)(y^2) + (x^5)(2yy')
d/dx(3x^3) = 9x^2
d/dx(-4) = 0
d/dx(1) = 0
and put it all together to get
y + xy' + 5x^4y^2 + 2x^5yy' + 9x^2 = 0
Now just solve for y' by collecting terms
y'(x+2x^5y) + (y + 5x^4y^2 + 9x^2) = 0
y' = -(y + 5x^4y^2 + 9x^2)/(x+2x^5y)
b) again, use the chain rule:
d/dx(ln y) = d/dy(ln y)*dy/dx = 1/y y'
1/y y' = -sin x
y' = -y sin x
thank for you so much steve!!!
i just wanted to double check.. for (a) it is y' = - (y+5x^4y^2+9x^2)/(x+2x^5y)
i was wondering if the last part of it (x+2x^5y) is it 2x^(5y) or 2x^5(y). im not sure if the "y" is in the exponent or not.
No, since we took the derivative of x^5*y^2 we get (5x^4)(y^2) + (x^5)(2yy')
That is, 2(x^5)(y)
ok so the final answer would b
a) y' = -(y + 5x^4y^2 + 9x^2)/(x+2x^5y)
and
b) y' = -y sin x
thank you SO much once again steve, my exam is in a couple of hours and this solution really helped me, im very thankful. thanks so much!!