I need some help with implicit differentiaiton.

[KFSKSS]

Hello. My problem is as follows: Suppose x^4+y^2+y-3=0. a) Compute dy/dx by implicit differentiation. b) What is dy/dx when x=1 and y=1? c) Solve for y in terms of x (by the quadratic formula) and compute dy/dx directly. Compare with your answer in part a).
I solved a) and b). a)=-4x^3/2y+1, and b)=-4/3. I'm stuck at c). This is what I've been doing: Using the quadratic formula to solve for y in x^4+y^2+y-3=0 gives y=(-1±√-4x^4+13)/2. Then applying the chain rule in the result of y gives -(4x^3)/√-4x^4+13. But it must give the same as a)= -4x^3/2y+1. Where am I failing at? How can I solve it?
Thank you for your time.

 

[Ray Vickson]

Hello. My problem is as follows: Suppose x^4+y^2+y-3=0. a) Compute dy/dx by implicit differentiation. b) What is dy/dx when x=1 and y=1? c) Solve for y in terms of x (by the quadratic formula) and compute dy/dx directly. Compare with your answer in part a).
I solved a) and b). a)=-4x^3/2y+1, and b)=-4/3. I'm stuck at c). This is what I've been doing: Using the quadratic formula to solve for y in x^4+y^2+y-3=0 gives y=(-1±√-4x^4+13)/2. Then applying the chain rule in the result of y gives -(4x^3)/√-4x^4+13. But it must give the same as a)= -4x^3/2y+1. Where am I failing at? How can I solve it?
Thank you for your time.

No: your solution for (a) is incorrect. You write
$$y' = -4 \frac{x^3}{2} y + 1,$$
but I get something quite different---unless, of course, you forgot to use parentheses!

 

[KFSKSS]

No: your solution for (a) is incorrect. You write
$$y' = -4 \frac{x^3}{2} y + 1,$$
but I get something quite different---unless, of course, you forgot to use parentheses!

Sorry I must have written it wrong. a)=(-4x^3)/2y+1. a) and b) are both correct since the book gives that answer too.

 

[Mark44]

Sorry I must have written it wrong. a)=(-4x^3)/2y+1. a) and b) are both correct since the book gives that answer too.

Your answer for a is also wrong, again due to missing parentheses.

What you wrote is $\frac{-4x^3}{2} y + 1$, according to the precedence rules of PEMDAS.
What you probably meant is almost certainly $\frac{-4x^3}{2y + 1}$

If you don't use LaTeX as I did, then what you wrote should have been -4x^3/(2y + 1), with parentheses around the two terms of the denominator. You could have parentheses around the numerator, but they aren't necessary.

LaTeX isn't hard to learn. We have a tutorial here: https://www.physicsforums.com/help/latexhelp/

 

[KFSKSS]

Your answer for a is also wrong, again due to missing parentheses.

What you wrote is $\frac{-4x^3}{2} y + 1$, according to the precedence rules of PEMDAS.
What you probably meant is almost certainly $\frac{-4x^3}{2y + 1}$

If you don't use LaTeX as I did, then what you wrote should have been -4x^3/(2y + 1), with parentheses around the two terms of the denominator. You could have parentheses around the numerator, but they aren't necessary.

LaTeX isn't hard to learn. We have a tutorial here: https://www.physicsforums.com/help/latexhelp/

Okay, thanks Mark44. I typed it wrong twice. I meant what you typed. Now, HOW CAN I SOLVE THE PROBLEM?

 

[epenguin]

You have told us what you meant, or rather you have told us that Mark has told us what you mean by solution (a ),

Now tell please in readable fashion what is your solution c including the steps getting there. Is that reasonable?