Question

How can we get dy/dx of sqrt(x+y)=1+x^2y^2 by using implicit differentiation?

Answer 1

by using the rules of derivatives that were taught during explicits

Answer 2

what would you say is the derivative of sqrt(u) ?

Answer 3

All you have to do is apply the chain rule. you know the derivative of y = dy/dx so anytime you are dealing with a derivative of y you multiply the derivative of the y related term by dy/dx. so for instance the derivative of xy^2 = y^2 + 2xy(dy/dx). Then, all you have to do is solve for dy/dx in terms of x and y.

Answer 4

nice and succinct

Answer 5

I see, thanks. After reading the textbook, I am still confused about some concept. For example, d/dx (7x+y)=7+dy/dx ? why should not equal 7+y*dy/dx?

Answer 6

well the derivative of a term such as a*u where a is a constant number is just a. so in this case we have the derivative of 1*u = 1 but we have to multiply it by the derivative of the inside which is dy/dx so we end up with d/dx (1*y) = 1*dy/dx or just dy/dx

Answer 7

my method is to ignore that this is spose to be wrtx and just take derivative like normal

D[7x+y]

D[x] + D[7y]

x' + 7 D[y]

x' + 7y'

now since x' wrtx = dx/dx = 1 ...

1 + 7y'

Answer 8

\[\sqrt{x+y}=1+x^2y^2\]
\[D[\sqrt{x+y}]=D[1+x^2y^2]\]
\[\frac{D[x+y]}{2\sqrt{x+y}}=D[1]+D[x^2y^2]\]
\[\frac{D[x]+D[y]}{2\sqrt{x+y}}=0+D[x^2]y^2+x^2D[y^2]\]
\[\frac{x'+y'}{2\sqrt{x+y}}=2x~x'~y^2+x^2~2y~y'\]\[x'=1\]
\[\frac{1+y'}{2\sqrt{x+y}}=2xy^2+2x^2y~y'\]
and algebra it into y' = .....

Answer 9

I see, thank you so much.