1. ## Implicit Differentiation/Related Rates

Question in manual:

Variables x and y are functions of t.

If (3x^2)y + 2x =2, and

dy= -4 when x=2 and y=-3,
dt

then find dx
sgggggggdt

I can't isolate the equation for x, which is why I have to use the implicit differentiation, but I'm not sure how to set up the solution. Any ideas?

2. ## Re: Implicit Differentiation/Related Rates

Originally Posted by Pewter12
Question in manual:

Variables x and y are functions of t.

If (3x^2)y + 2x =2, and

dy= -4 when x=2 and y=-3,
dt

then find dx
sgggggggdt

I can't isolate the equation for x, which is why I have to use the implicit differentiation, but I'm not sure how to set up the solution. Any ideas?
you do not need to isolate x ... you need to isolate dx/dt after you take the time derivative.

3. ## Re: Implicit Differentiation/Related Rates

As a start, let's what your result is for the implicit differentiation. (This should be differentiation with respect to t.)

4. ## Re: Implicit Differentiation/Related Rates

Thank-you for your replies. My question was not very clear. I understand the question is asking for dx/dt, and that taking the derivative of t comes first. Unfortunately I'm entirely confused as to how to the x=2 and y=-3 fit in, because the dx/dt format and derivatives in general are very new to me.

5. ## Re: Implicit Differentiation/Related Rates

$\displaystyle \frac{d}{dt}\left(3x^2y + 2x = 2\right)$

note that the product rule is required for the first term ...

$\displaystyle 3x^2 \cdot \frac{dy}{dt} + y \cdot 6x \cdot \frac{dx}{dt} + 2\frac{dx}{dt} = 0$

... isolate $\displaystyle \frac{dx}{dt}$ , sub in your given values and finish it.

6. ## Re: Implicit Differentiation/Related Rates

Thank-you, that is the clarification I was hoping for.