I have to find y' if ln(xy) = x + y.
Where do I start? I'm not sure what to do with that ln(xy)
What you do with ln(xy) is just separate it into 2 pieces: ln(x) + ln(y)
ln(x) + ln(y) = 1 + y'
1/x + y'/y = 1 + y'
1/x - 1 = y' (1 - y)
use the chain rule on the left:
d/dx(ln(u)) = 1/u du/dx
u = xy, so
du/dx = dx/dx*y + x*dy/dx = y + xy'
now plug and chug:
1/(xy) * (y + xy') = 1 + y'
y/xy + xy'/xy = 1 + y'
1/x + y'/y = 1 + y'
y'(1/y - 1) = 1 - 1/x
or
y' * (1-y)/y = (x-1)/x
y' = y(x-1)/[x(1-y)]
So then that could be moved around, and it would be (1-x)/(x-xy) ?
Oh geez two different answers.
That's right. Do you see where Raf left out the 1/y?
ln(x) + ln(y) = 1 + y'
1/x + y'/y = 1 + y'
1/x - 1 = y' (1 - 1/y)
tnx for the correction Steve, my bad Sarah
To find y', the derivative of y with respect to x, you can start by taking the derivative of both sides of the equation with respect to x. However, before doing that, you need to apply the product rule to differentiate the term ln(xy).
The product rule states that for two functions u(x) and v(x), the derivative of their product is given by:
(d/dx) [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)
So, let's differentiate ln(xy) using the product rule:
(d/dx) [ln(xy)] = (d/dx) ln(uv) = (u'/u + v'/v) * uv
In this case, u(x) = x and v(x) = y. Therefore, u'(x) = 1 and v'(x) = y'.
Applying the product rule to the left-hand side of the equation, we have:
(d/dx) [ln(xy)] = (1/x) * xy + ln(xy) * y'
Simplifying this, we get:
1/x * xy + ln(xy) * y' = x + y
Now, we can solve for y' by isolating it on one side:
ln(xy) * y' = x + y - 1/x * xy
Finally, to obtain y', divide both sides of the equation by ln(xy):
y' = (x + y - 1/x * xy) / ln(xy)
So, the value of y' can be calculated using the expression above.