Question

find y' using implicit differentiation of the following: (2x^2) + x + xy=1

Answer 1

A good way to remember is that implicit differentiation refers to the fact that y is an implied function of x, so according to chain rule:

\[ \frac{d}{dx}(y(x))=y' \]

Whenever you derive y, the chain rule kicks out a y' so , lets work through the differentiation

\[ \frac{d}{dx}(2x^2) + \frac{d}{dx}(x) + \frac{d}{dx}(xy) = \frac{d}{dx}(1) \]

Most of those derivatives should be obvious, but lets focus on:

\[ \frac{d}{dx}(xy) \]

This is a product rule so we get (remember that d/dx (y ) = y'

\[ \frac{d}{dx}(xy) = y + y'x \]

Now your derivative becomes:

\[ 4x + 1 + x +xy'=0\]

I'll leave it at that, since now you can use algebra to solve for y'

Answer 2

awesome! although I don't seem to follow the y+y'x, I do understand how you got the final answer, but now how would I solve explicitly for y to get y' in terms of x?? or do I already have that?

Answer 3

would y'=(-4x-1-x)/x ?

Answer 4

Whoops, there is a typo in my answer:

But you do not have to solve for y, the derivative itself is a function of x and y.

Yes, the answer would be that, with one change due to my typo:

\[ 4x + 1 + y + xy' = 0 \]

\[ y' = \frac{-4x-1-y}{x}\]

To get the xy portion, here's a more indepth explanation. According to product rule:

\[ \frac{d}{dx}(xy) = \frac{d}{dx}(x)y + \frac{d}{dx}(y)x \]

Then
\[ \frac{d}{dx}(x) = 1 \]

\[ \frac{d}{dx}(y) = y' \]

\[ \frac{d}{dx}(xy) = y + y'x \]

Answer 5

awesome this makes sense ! would you mind helping me with one more? i have cosx+sqrty=5 the root y is messing me up!

Answer 6

Sure, it gets a bit trickier here, lets look at that derivative by itself (technically a double chain rule will occur)

\[ \frac{d}{dx}(\sqrt{y})= \frac{d}{dx}((y)^{1/2}) \]

Here you first have to take the derivative of the power and then the derivative of y which will be y', so from the chain rule we get:

\[ \frac{d}{dx}(\sqrt{y})= \frac{d}{dx}((y)^{1/2}) = \frac{1}{2(y^{-1/2})}(y')= \frac{y'}{2(y^{-1/2})}\]

The y' was a chain rule, they way I like to do it is think of y as just another variable. If you had x^1/2, what would its derivative be? What ever that is, just make sure to apply the chain rule and kick out a y' afterwards. Try using that to finish solving the problem and see what answer you get.

Answer 7

great! thank you so much!

Answer 8

Jeeze, its getting late, once again a typo, I jumped the gun it. The derivative, if the y is in the denominator, should be y^(1/2)

Answer 9

oh okay! hahaha thanks again!