[Implicit Differentiation]
(5x-y)^4 + 2y^3 = 258 evaluate y' at the point (1,1)
Check that the point (1,1) satisfies the given equation, i.e. it is on the curve.
Use implicit differentiation to differentiate each term and obtain an equation containing x, y and y'.
Solve for y'.
You should get
y'=20(5x-y)³/(4(5x-y)³-6y²)
Substitute (1,1) into the above equation to get y'(1,1)=128/25.
Check my calculations.
Thanks. Correct.
You're welcome!
To evaluate y' at the point (1, 1) using implicit differentiation, follow these steps:
Step 1: Differentiate both sides of the equation with respect to x.
For the left side, we can use the chain rule since the term (5x-y) is raised to the power of 4. The derivative of (5x-y) with respect to x is 5 - y'. Applying the chain rule, we multiply this by the derivative of (5x-y), which is 5.
For the right side, the derivative of 258 (a constant) with respect to x is 0.
Step 2: Simplify and solve for y'.
Using the differentiation rules, we have:
5(5x-y)^3(5) - 3(2y^2)(y') = 0
25(5x-y)^3 = 6y^2y'
Step 3: Plug in the given point (1, 1) to find y'.
Substitute x = 1 and y = 1 into the equation:
25(5(1)-1)^3 = 6(1)^2y'
25(5-1)^3 = 6y'
25(4)^3 = 6y'
25(64) = 6y'
1600 = 6y'
Simplify the equation further:
Divide both sides by 6:
1600/6 = y'
266.67 ≈ y'
Therefore, y' at the point (1, 1) is approximately 266.67.