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

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

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

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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/ – if x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1.

y′=−2y+3y2

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

y′′=(y+3y2)(0)−(−2)(y+3y2)′(y+3y2)2

(y+3y2)′=(1)(dydx)+6y(dydx)=−2y+3y2+−12yy+3y2=−2−12yy+3y2

y′′=0−(−2)(−2−12yy+3y2)(y+3y2)2=−4+24yy+3y2(y+3y2)2=−4+24yy+3y2∗1(y+3y2)2

y′′=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!

y′′′=(y+3y2)3(24)(dydx)−(4+24y)((y+3y2)3)′(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 f′(x)=3x2 If we then let g(x)=y+3y2 then g′(x)=1+6y leaving us with…

3(y+3y2)2∗(1+6y)

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

y′′′=(y+3y2)3(24)(dydx)−(4+24y)(3(y+3y2)2∗(1+6y))(dydx)(y+3y2)6

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

y′′′=(y+3y2)2(y+3y2)(24)(dydx)−(4+24y)(3(1+6y))(dydx))(y+3y2)6

cancelling out for…

y′′′=(y+3y2)(24)(dydx)−(4+24y)(3(1+6y))(dydx))(y+3y2)4

y′′′=24y(1+3y)(dydx)−12(36y2+12y+1)(dydx)(y+3y2)4

y′′′=−48y(1+3y)y+3y2+24(36y2+12y+1)y+3y2(y+3y2)4=−48y(1+3y)+24(36y2+12y+1)y+3y2(y+3y2)4

Then finally…

y′′′=−48y(1+3y)+24(36y2+12y+1)y+3y2∗1(y+3y2)4

y′′′=−48y(1+3y)+24(36y2+12y+1)(y+3y2)5

homework.study.com/ – if x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1.

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 y′′′ 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 ′s′ w.r.t ′w′

is given by; s

dsdw=−(∂f∂w∂f∂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)dpdx−p(x)dqdx(q(x))2

https://brainly.in – if x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1.

Given

x² + xy + y³ = 1

x = 1

To find

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

Solution

If x² + xy + y³ = 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,

x² + xy + y³ = 1

Substitute x = 1,

1² + (1)y + y³ = 1

1 + y + y³ = 1

y + y³ = 1 – 1

[y + y³ = 0]

y ( 1 + y²) = 0

1 + y² = 0

y² = -1

y =

Final answer – if x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1.

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

Answer

y = -1

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Explanation – if x2 + xy + y3 = 1, find the value of y”’ at the point where x = 1.

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

1×2+1×y+y×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

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