Evaluate y' at (2,1) [Implicit differentiation]
5xy-3x-4=0
I got 3/10 and 2/10 but both were wrong.
What do you mean by "both" ?
(2,1) is the single point where x=2 and y=1.
Using implicit differentiation to get dy/dx, consider y to be a function of x and differentiate both sides. This results in
5x dy/dx + 5y -4 = 0
When x=2 and y=1,
10 dy/dx + 5 -4 = 0
dy/dx = -1/10
By both, I mean I got my answer to be 3/10 and 2/10 but they weren't correct.
I tried -1/10 and it was wrong.
The answer is -2/10.
5x dy/dx + 5y -4 = 0
There was a typo. the -4 should read -3.
You will get -2/10 or -1/5.
To find the value of y' (the derivative of y with respect to x) at a specific point (2,1) using implicit differentiation, follow these steps:
1. Start by differentiating both sides of the equation with respect to x.
The equation is: 5xy - 3x - 4 = 0
Differentiating both sides: d/dx(5xy - 3x - 4) = d/dx(0)
2. Apply the product rule when differentiating the term 5xy:
d/dx(5xy) - d/dx(3x) - d/dx(4) = 0
3. Find the derivatives of each individual term:
For 5xy, the derivative with respect to x is: 5y + 5x(dy/dx)
For -3x, the derivative with respect to x is: -3
For 4 (a constant), the derivative with respect to x is: 0
4. Substitute the values back into the differentiated equation:
5y + 5x(dy/dx) - 3 - 0 = 0
5. Rearrange the equation and solve for dy/dx:
5x(dy/dx) + 5y = 3
5x(dy/dx) = 3 - 5y
dy/dx = (3 - 5y) / 5x
6. Plug in the point (2,1) to find the value of y' at that specific point:
dy/dx = (3 - 5(1)) / (5(2))
dy/dx = (3 - 5) / 10
dy/dx = -2/10
dy/dx = -1/5
Therefore, the value of y' at the point (2,1) is -1/5.