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# if x2+xy+y3=1, find the value of ‘y’ at the point where x=1

• First of all, we want to talk about how to find the value of “if x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1″. Here we have so many websites to give an excellent answer to this question. Like chegg.com/, math.stackexchange.com/, homework.study.com/ and also etc. All these websites are trying to provide perfect and also proper solutions for students.

## Question – if x2+xy+y3=1, find the value of y at the point where x =1

When I search for â€œif x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1â€ I saw this below one,

They are not providing the exact answer for this “if x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1″ question. No problem, we have so many other websites also. So next, we will look into math.stackexchange.com/. Let’s see what they are trying to say.

## math.stackexchange.com

yâ€²=âˆ’2y+3y2

Then trying to take the second derivative using the quotient rule.

• yâ€²â€²=(y+3y2)(0)âˆ’(âˆ’2)(y+3y2)â€²(y+3y2)2
• (y+3y2)â€²=(1)(dydx)+6y(dydx)=âˆ’2y+3y2+âˆ’12yy+3y2=âˆ’2âˆ’12yy+3y2
• yâ€²â€²=0âˆ’(âˆ’2)(âˆ’2âˆ’12yy+3y2)(y+3y2)2=âˆ’4+24yy+3y2(y+3y2)2=âˆ’4+24yy+3y2âˆ—1(y+3y2)2
• yâ€²â€²=4+24y(y+3y2)3

At this point, I’m not sure if this is right. But I went and tried for the third with a more quotient rule!

yâ€²â€²â€²=(y+3y2)3(24)(dydx)âˆ’(4+24y)((y+3y2)3)â€²(y+3y2)6

Before I can finish that problem, I need the derivative of (y+3y2)3… for this, I need to use the chain Rule.

if f(x)=x3 then fâ€²(x)=3x2 If we then let g(x)=y+3y2 then gâ€²(x)=1+6y leaving us with…

3(y+3y2)2âˆ—(1+6y)

Plugging that back into the rest of the problem, we get…

yâ€²â€²â€²=(y+3y2)3(24)(dydx)âˆ’(4+24y)(3(y+3y2)2âˆ—(1+6y))(dydx)(y+3y2)6

factoring out a (y+3y2)2, I get

yâ€²â€²â€²=(y+3y2)2(y+3y2)(24)(dydx)âˆ’(4+24y)(3(1+6y))(dydx))(y+3y2)6

cancelling out for…

• yâ€²â€²â€²=(y+3y2)(24)(dydx)âˆ’(4+24y)(3(1+6y))(dydx))(y+3y2)4
• yâ€²â€²â€²=24y(1+3y)(dydx)âˆ’12(36y2+12y+1)(dydx)(y+3y2)4
• yâ€²â€²â€²=âˆ’48y(1+3y)y+3y2+24(36y2+12y+1)y+3y2(y+3y2)4=âˆ’48y(1+3y)+24(36y2+12y+1)y+3y2(y+3y2)4

Then finally…

yâ€²â€²â€²=âˆ’48y(1+3y)+24(36y2+12y+1)y+3y2âˆ—1(y+3y2)4

yâ€²â€²â€²=âˆ’48y(1+3y)+24(36y2+12y+1)(y+3y2)5

## homework.study.com

According to homework.study.com, to solve the problem, “if x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1″.

## Question:

If x2+xy+y3=1, find the value of yâ€²â€²â€² at the point where x = 1. Use Implicit differentiation.

## Implicit Differentiation and Quotient Rule

Any function f(w,s)=c is called an implicit function for that derivative of â€²sâ€² w.r.t â€²wâ€²

is given by; s

dsdw=âˆ’(âˆ‚fâˆ‚wâˆ‚fâˆ‚s)

Let, function F(x)

in the form of fractions such that;

F(x)=p(x)q(x)

Derivative of F(x)

is given by;

dFdx=ddx(p(x)q(x))

dFdx=q(x)dpdxâˆ’p(x)dqdx(q(x))2

## https://brainly.in

### Given

xÂ² + xy + yÂ³ = 1

x = 1

### To find

The value of y at the point where x = 1.

### Solution

If xÂ² + xy + yÂ³ = 1

We can find the value of y at the point where x = 1 by substituting the value of x in the given equation.

So,

xÂ² + xy + yÂ³ = 1

Substitute x = 1,

1Â² + (1)y + yÂ³ = 1

1 + y + yÂ³ = 1

y + yÂ³ = 1 – 1

[y + yÂ³ = 0]

y ( 1 + yÂ²) = 0

1 + yÂ² = 0

yÂ² = -1

y =

The value of y at the point where x = 1 is

y = -1

4

### Explanation

x2+xy+y3=1. (if x is 1)

1Ã—2+1Ã—y+yÃ—3=1

2+y+3y=1. (put 2 to another side)

y+3y=1-2. add 3y and y,subtract1-2 )

4y=-1

y=-1 or y= (4)power -1

4

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