Finding y prime using implicit differentation?

Question

How do i fing y prime using implicit differentation? the problem is: xlny^2=e^2y

Answer 1

Spaces and parentheses are a virtue. Also, what you have literally written is x (ln y)² = e² y, but somehow I suspect you meant to write x ln (y²) = e^(2y). If you actually meant what you wrote, my mistake, but I'll go with what I think you mean based on experience with interpreting problems on Y!A.

x ln (y²) = e^(2y)

First, we can actually simplify the expression on the left:

2x ln y = e^(2y)

Now, taking the derivative of both sides:

d(2x ln y)/dx = d(e^(2y))/dx

Using the product rule on the left and the chain rule on the right:

d(2x)/dx ln y + 2x d(ln y)/dx = e^(2y) d(2y)/dx

Simplifying the derivatives, again using the chain rule:

2 ln y + 2x y'/y = 2y' e^(2y)

Dividing both sides by 2, and subtracting xy'/y:

ln y = y'e^(2y) - xy'/y

Factoring the right:

ln y = (e^(2y) - x/y)y'

Dividing by e^(2y) - x/y:

y' = ln y/(e^(2y) - x/y)

And we are done.

Answer 2

It's sort of hard to type things here, isn't it? I hope I'm right in assuming that the left side is xln(y^2), meaning y squared then natural log. If not hollar and I will change this! ^_^

With implicit differentiation, treat y not as just a variable, but as a function of x, only you don't know what the function is. You'll need to use the product rule whenever you have an x and a y (or functions of them) multiplied by each other. And whenever you take the derivative of a function of y, you need to use the chain rule.

So taking the derivative of the left side...
xln(y^2)
= 2xlny (You can pull out the exponent)
Now use the product rule: first x second' + second x first' to get:
2x(1/y)(dy/dx) + 2ln(y)
(Note that taking the derivative of y means you're taking the derivative of a function of x, so when you do y', remember this rule: if F(x) = f(g(x)), then F'(x) = f '(g(x)) g '(x). You don't know what the "inner function" is though, so for its derivative, you use dy/dx.)

Now for the right side:
e^(2y)
Replace 2y with u.
e^u, where u is a function.
Take the derivative to get:
u'e^u (You have to take the derivative of u, also, since u is a function.)
Since u = 2y, u' = 2y' = 2(dy/dx)
So...
u'e^u = 2(dy/dx)e^(2y)

So the derivative of the function ends up being:
2x(1/y)(dy/dx) + 2ln(y) = 2(dy/dx)e^(2y)
Now solve for dy/dx (which is y')
2x(1/y)(dy/dx) - 2(dy/dx)e^(2y) = -2ln(y)
(dy/dx)[2x(1/y)- 2e^(2y)] = -2ln(y) .......[note that you can factor out dy/dx]
dy/dx = y' = -2ln(y)/[2x(1/y)- 2e^(2y)]

Answer 3

take care of 2xy as a product, so (2xy)' = 2xy' + y(2). So the finished shebang is 2xy' + 2y + y' = 0 settle on for y' y'(2x + a million) = –2y y' = 2y / (2x + a million) So on the indicated aspect, y'(–a million, 2) = 2(2)/(2(–a million)+a million = 4/(–a million) = –4

Answer 4

xlny^2=e^2y
lny^2+x/y^2 *2yy'= e^2
lny^2 + 2xy'/y = e^2
ylny^2 +2xy' = ye^2
y' = (ye^2 -ylny^2)/2x