Find the slope of the tangent line to the curve below at the point (1,6)

Question

√(4x+2y) + √(4xy) = 8.8989794855664

Everything in the parentheses is under the square root symbol.

At the point (1,6) what is the slope of the tangent line to the curve?

Leave you answer as a negative decimal number.

Answer 1

So to do this we will need to perform implicit differentiation. This just means that there are "y"s in our equation that we need to care about.

When we take an implicit derivative, we need to consider the fact that "y" or "f(x)" is some function of "x", but we don't necessarily know what it is. We could perhaps solve for "y," but that is a lot of work, and sometimes isn't even possible! So instead, if we can go by the following; "y" contains some "x"s, so we can use the chain rule to allow for its derivative to be taken. So any time we see a "y", it's derivative will be, just as the chain rule says, the derivative of your function of "y", multiplied the derivative of the inside of "y", which we just call dy/dx or y'.

Now let's take the derivative your problem. We can split it into three terms

First:

√(4x+2y) = (4x+2y)^1/2 : power rule/chain rule

1/2(4x + 2y)^-1/2 * (4 + 2y')

Second:

√(4xy) = (4xy)^1/2 : power rule/product rule/chain rule

1/2(4xy)^-1/2 * (4y + 4y')

Third:

8.8989794855664

This simply derives to 0

So all together, we have:

1/2(4x + 2y)^-1/2 * (4 + 2y') + 1/2(4xy)^-1/2 * (4y + 4y') = 0

Now from here, you simply plug in 1 for every x you see and 6 for every y you see. Then, you simply solve for y'

1/2(4(1) + 2(6))^-1/2 * (4 + 2y') + 1/2(4(1)(6))^-1/2 * (4(6) + 4y') = 0

(1/8)(4 + 2y') + (1/(4√6))(24 +4y') = 0

To simplify, let's call 1/8 a and 1/(4√6) b

a(4 + 2y') + b(24 +4y') = 0

4a + 24b + 2ay' + 4by' = 0

(2a + 4b)y' = -(4a + 24b)

Since your answer should be a negative decimal, just replace a and b with 1/8 and 1/(4√6) respectively

2(1/8) + 4(1/(4√6)) = -(2(1/8) + 4(1/(4√6)))

0.6582482905y' ≈ -2.949489743

y' ≈ -4.480816412

And that should be your answer