Calculus I quiz prep help

Question

For each of the following curves, find a general expression for dy/dx . Then evaluate the derivative at the point indicated and write the equation of the tangent line to the curve at the point

a) x^2 − y^2 = 1 at (√ 5, 2)

b) 2x^2 − xy + 3y^2 = 18 at (3, 1)

c) 2x^3 + 2y^3 = 9xy at (2, 1).

Answer 1

If you had these equations in the form "y = (some function of x)" then you could just use the regular rules you've learned for differentiation (power rule, product rule, etc). But unfortunately, they are not in that form. There are a couple of ways of approaching this. 1) Use algebra to put them into the form y = (some function of x), 2) use implicit differentiation. They obviously are wanting you to focus on the later.

I'll do one of these problems for you to show you how it's done - how about problem c) 2x³ + 2y³ = 9xy

Step 1) apply the differentiation operator to both sides of the equation:

d(2x³ + 2y³)/dx = d(9xy)/dx

Step 2) Differentiate with respect to x remembering that "y" is some complex function in terms of x (that you do not know) and requires you to use the chain rule on it. So the derivative of 2x³ with respect to x is 6x² (simple power rule). Then, the derivative of 2y³ with respect to x is 6y² * dy/dx (simple power rule BUT you need to take the derivative of what's inside by the chain rule and you don't know what it is except that it can be written as dy/dx). Then, on the other side, you have two things being multiplied so you need to use the product rule (if f(x) = u*v then f '(x) = u'v + uv') so to find the derivative of 9xy with respect to x , let u = 9x and v = y then u' = 9 and v' = dy/dx so the derivative of 9xy with respect to x is 9y + 9x * dy/dx. Putting that all together gives us:

6x² + 6y²(dy/dx) = 9y + 9x(dy/dx)

We now need to solve for dy/dx because that's what we are looking for. Bringing all the dy/dx 's together on one side of the equation and everything else on the other gives:

6y²(dy/dx) - 9x(dy/dx) = 9y - 6x²

then factoring out dy/dx we get:

dy/dx(6y² - 9x) = 9y - 6x²

Now, divide both sides by (6y² - 9x) gives:

dy/dx = (9y - 6x²)/(6y² - 9x)

That's the general expression for dy/dx. To find the derivative at the point, just plug in x = 2 and y = 1 (the point is (2, 1) and we get dy/dx = (9(1) - 6(2)²)/(6(1)² - 9(2)) = -15/-12 = 15/12

So 15/12 is the slope of the tangent line at (2, 1). To find the equation of the line, use the point-slope form [(y - y₁) = m(x - x₁)] where (x₁, y₁) is (2, 1) and you get y - 1 = 15/12(x - 2)

You do the others all the same way.