Coefficient of the tangent/slope

[Loki123]

My idea was just to bring y to the LHS and calculate derivative since y' = m, however that's not possible. So I googled and found some people telling to find critical points, but I did not get it in the end...
What do I do

 

[BigBeachBanana]

I think this is poor English. By "the coefficient of the direction of the tangent to the curve..." Do you mean the slope of the tangent line?

 

[Loki123]

I think this is poor English. By "the coefficient of the direction of the tangent to the curve..." Do you mean the slope of the tangent line?

yes, i translated it literally, but i put the real meaning in the title. i remembered it too late.

 

[BigBeachBanana]

yes, i translated it literally, but i put the real meaning in the title. i remembered it too late.

For a problem like this, implicit differentiation makes it really simple. Have you learned it since the angle of the intersection thread? If not, here's a link.

Calculus I - Implicit Differentiation

In this section we will discuss implicit differentiation. Not every function can be explicitly written in terms of the independent variable, e.g. y = f(x) and yet we will still need to know what f'(x) is. Implicit differentiation will allow us to find the derivative in these cases. Knowing...

tutorial.math.lamar.edu

 

[pka]

My idea was just to bring y to the LHS and calculate derivative since y' = m, however that's not possible. So I googled and found some people telling to find critical points, but I did not get it in the end...
What do I doView attachment 32562

Please note that we want the tangent to [imath]2y=1+xy^2[/imath] at [imath](1,1)[/imath] which is point on the curve.
Rewrite the curve [imath]xy^3-2y+1=0[/imath] thin do implicit differentiation :
[imath]y^3+3xy^2y^{\prime}-2y^{\prime}=0\text{ or }y^{\prime}=\dfrac{-y^3}{3xy^2-2}[/imath]
What is [imath]y^{\prime}(1,1)=~?[/imath][imath][/imath]

 

[JeffM]

I too am confused by the language of the problem.

If it is asking for the slope of the tangent line at point (1, 1), that makes sense.

It looks as though you tried to do implicit differentiation, but made at least three mistakes.

[math]xy^3 - 2y + 1 = 0 \implies y^3 + 3xy^2y' - 2y' = 0 \implies\\ y^3 = 2y' - 3xy^2y' = y'(2 - 3x^2) \implies y' = \dfrac{y^3}{2 - 3x^2}.[/math]
When differentiating [imath]xy^3[/imath], you forgot both the product rule and the chain rule. When differentiating [imath]2y[/imath], you forgot the chain rule. You are differentiating with respect to x, not with respect to y, so you need the chain rule.

 

[Loki123]

For a problem like this, implicit differentiation makes it really simple. Have you learned it since the angle of the intersection thread? If not, here's a link.

Calculus I - Implicit Differentiation

In this section we will discuss implicit differentiation. Not every function can be explicitly written in terms of the independent variable, e.g. y = f(x) and yet we will still need to know what f'(x) is. Implicit differentiation will allow us to find the derivative in these cases. Knowing...

tutorial.math.lamar.edu

What do you think, am I correct?

 

[BigBeachBanana]

What do you think, am I correct?
View attachment 32579

yes, correct.

 

[Loki123]

yes, correct.

thanks