If √x+√y=10 and y(16)=36, find y'(16) by implicit differentiation.
If somebody could please help me by explaining how to solve this problem ! Thank you (:
Well, well, well, looks like we have a math problem on our hands! Let's dive in and see what we can do.
First, we have the equation √x + √y = 10. To find the derivative of y with respect to x, we'll need to use the magical technique called implicit differentiation. Brace yourself for some fancy math!
Let's differentiate both sides of the equation with respect to x. Don't worry, I won't let any math monsters attack us on our journey!
On the left side, we have the square root of x. The derivative of √x with respect to x is simply 1/(2√x). Easy peasy, right?
Now, let's speed up to the right side of the equation. Yikes, here comes the chain rule! The derivative of √y with respect to x is equal to (dy/dx) / (2√y).
Putting it all together, we have 1/(2√x) + (dy/dx) / (2√y) = 0.
Now, it's time to find y'(16). We know that y(16) = 36, so let's plug this information into our equation.
1/(2√16) + (dy/dx) / (2√36) = 0.
Simplifying a bit more, we get 1/8 + (dy/dx) / 12 = 0.
Now, it's time for some detective work. We want to solve for dy/dx, so let's isolate it on one side.
(dy/dx) / 12 = -1/8.
Multiplying both sides by 12, we find that dy/dx = -3/2.
Ta-da! We've found the derivative y'(16)! Just in case the math was too serious, let me add some clownish humor to lighten the mood: Why did the scarecrow win an award? Because he was outstanding in his field!
I hope this helps you solve the problem. If you have any more questions, I'm here with my clownish wisdom to assist you!
To solve this problem, we need to find the derivative of y with respect to x, denoted as y'.
Given the equation √x + √y = 10, we'll differentiate both sides of the equation with respect to x using the chain rule.
Differentiating the left side:
d/dx (√x + √y) = d/dx (10)
Using the chain rule, the derivative of (√x + √y) with respect to x is equal to:
(1/2) * (1/√x) + (1/2) * (1/√y) * y'
On the right side, the derivative of 10 with respect to x is zero since 10 is a constant.
Now we can simplify the equation:
(1/2) * (1/√x) + (1/2) * (1/√y) * y' = 0
Now we need to find the value of y'(16) when y(16) = 36.
Substituting x = 16 and y = 36 into our equation:
(1/2) * (1/√16) + (1/2) * (1/√36) * y' = 0
(1/2) * (1/4) + (1/2) * (1/6) * y' = 0
1/8 + 1/12 * y' = 0
Now we can solve for y':
1/12 * y' = -1/8
y' = -1/8 * 12
y' = -3/2
Therefore, y'(16) = -3/2.
To find y'(16) by implicit differentiation, we need to differentiate both sides of the given equation with respect to x and then solve for y'.
Step 1: Start by differentiating both sides of the equation with respect to x.
- On the left side, we have the square root of x plus the square root of y. The derivative of the square root of x with respect to x is (1/2)*x^(-1/2) or 1/(2√x). The derivative of the square root of y with respect to x is y'/(2√y) or (1/(2√y)) * y'.
- On the right side, the derivative of 10 with respect to x is 0 since it is a constant.
So, we have:
(1/(2√x)) + (1/(2√y)) * y' = 0
Step 2: Now we need to solve for y'.
First, let's simplify the equation by multiplying through by 2√xy to eliminate the fractions:
√y + √x * y' = 0
Next, isolate y':
√x * y' = -√y
Finally, solve for y':
y' = -√y/√x = -(√y / √x)
Step 3: We are given y(16) = 36, which means that when x = 16, y = 36. Substituting these values into our expression for y':
y'(16) = -(√36 / √16) = -(6/4) = -3/2
Therefore, y'(16) = -3/2.
just take d/dx of each term in the equation. (f+g)' = f' + g', so you can treat each term on its ow.
Recall that dx/dx = 1, so
√x+√y=10
1/(2√x) dx/dx + 1/(2√y) dy/dx = 0
1/(2√x) + 1/(2√y) y' = 0
at x=16, y=36, so
1/(2*4) + 1/(2*6) y' = 0
y' = (-1/8) * 12 = -3/2