# Thread: Calculus problem(I have attached the picture of the question)

2. ## Re: Calculus problem(I have attached the picture of the question)

The question is asking you to use implicit differentiation. Have you covered implicit differentiation? For part (b), you should plug in $x=1$ to the equation to find all values for $y$ that satisfy the equation (you will have an easily solvable quadratic equation). Once you find the two values of $y$ that satisfy the equation, you will have two points to find the derivative.

Please identify specifically where you are having trouble with this problem. I can easily come up with a solution, but it may not address your specific area of concern.

3. ## Re: Calculus problem(I have attached the picture of the question)

The equation is $\displaystyle 3x- 2y^2e^{x- 1}= 2$ and you want to determine $\displaystyle \frac{dy}{dx}$. Do you know what the derivative of "3x", with respect to x, is? What about the derivative of "2"? To differentiate the term $\displaystyle 2y^2e^{x- 1}$ you can use the "product rule" first. $\displaystyle (2y^2e^{x- 1})'= 2y^2(e^{x- 1})'+ 2(y^2)'(e^{x- 1})$. To differentiate $\displaystyle y^2$, with respect to x, use the chain rule.