Did I get this right?? (Implicit Diferentiation)

[elindow]

I am doing Implicit Diferentials and I had this problem and I wanted to know if I got it right. (It actually starts out square root but I don't know how to do that here)

√(x+y) = 1 + x²y²

This is how I worked it:

(x=y)^1/2 = 1 + x²y²

1/2(x+y)^-1/2 = 0 + (x²2y + y²2x * y')

-2x²y - y'(-2xy²) + 1/2(x+y)^-1/2 = 0

y'(-2xy²) = (2x²y)/(1/2(x+y)^1/2)

(2x²y)(-2xy²) = y’
-------------
1/2(x+y)^1/2

Could someone please tell me if I did this right??

Thanks
Erik

 

[leijonah]

Im no math expert but I think you forgot to differentiate a part

YOu need to have this step
1/2(x+y)^-1/2 * (1+yprime)=0+2xy^2+(x^2*2y*yprime)
the first part is chain rule application, then you group the yprimes by first getting rid of the 1+yprime, by expanding, then solve for yprime by factoring, etc....

 

[soroban]

Hello, elindow!

I am doing Implicit Differentiation.

√(x+y) = 1 + x²y²

This is how I worked it:

(x+y)^1/2 = 1 + x²y²

1/2(x+y)^-1/2 = 0 + (x²2y + y²2x·y')
. . . . . . . . . ↑ . . . . . . ↑ . . . . . . ↑
. . . . . . . . . . . . . . errors!

As leijonah pointed out, you need the Chain Rule for the first term.

You differentiated the exponent correclY: .½(x + y)^-½

. . then you must "go inside" and differentiate: .1 + y'


On the right side, you used the Product Rule incorrectly.

We want the derivative of x²y².

Product Rule: . x²·(der. of y²) + y²·(der. of x²)

. . . . . . . . . . . . . x²·(2y.y') . . .+ . . .y²·(2x)

 

[tkhunny]

I have to go with leijonah on this one. (Finnish Lion?)

You may wish to practice on the chain rule and product rule a bit more. You missed a piece of the chain rule on the left-hand since and I guess you just confused yourself on the product rule on the right-hand side.