Slope of a curve confusion....

[calculusdunce]

I know that I'm supposed to apply this rule here:

For a function, y=f(x), the derivative, y’, evaluated at x=c can be represented in symbolic form as [MATH]\frac{dy}{dx} |c[/MATH].

 

[Deleted member 4993]

View attachment 27981
I know that I'm supposed to apply this rule here:

For a function, y=f(x), the derivative, y’, evaluated at x=c can be represented in symbolic form as [MATH]\frac{dy}{dx} |c[/MATH].

You are saying:

For a function, y=f(x), the derivative, y’, evaluated at x=c can be represented in symbolic form as \(\displaystyle \frac{dy}{dx}|_c\)

You are correct.

So apply it - and show us what you get, and continue....

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[HallsofIvy]

If you are concerned that you cannot write y as an expicit function of x, use "implicit" differentiation.

\(\displaystyle y^2- xy^3= 10\).

The derivative of \(\displaystyle y^2\) with respect to x is \(\displaystyle 2y\frac{dy}{dx}\).

The derivative of \(\displaystyle xy^3\) is \(\displaystyle y^3+ 3xy^2\frac{dy}{dx}\).

And, of course, the derivative of "10" is 0.

So the derivative of \(\displaystyle y^2- xy^3= 10\) is

\(\displaystyle 2y\frac{dy}{dx}- y^3 - 3xy^2\frac{dy}{dx}= 0\)

so \(\displaystyle (2y- 3xy^2)\frac{dy}{dx}= y^3\). Solve that for \(\displaystyle \frac{dy}{dx}\).

 

[jonah2.0]

Beer inspired diagram follows.

I know that I'm supposed to apply this rule here:

For a function, y=f(x), the derivative, y’, evaluated at x=c can be represented in symbolic form as [MATH]\frac{dy}{dx} |c[/MATH].