The equation of a curve is y =√(8x − x2). Find an expression for dy/dx and the coordinates of the stationary point on the curve,

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Hi, saj, For this, we can use the chain rule to find the derivative. So: dy/dx [(8x − x^2)^(1/2)] = 1/2 (8x - x^2)^(-1/2) * (8 - 2x) Simplifying a bit: dy/dx = (4-x)/(8x - x^2)^(-1/2) For the stantionary point, we set the derivative = 0. So, we are solving the equation: 0 = (4-x)/(8x - x^2)^(-1/2) For this to be true, the top has to be 0. So: 4-x = 0 x = 4 To find the y coordinate, we plug this back into the original equation: y = √(8*4 − 4^2) = √(32 - 16) = 4 So, the stationary point is (4,4) Good luck, saj. I hope this helps. Till Then, Steve
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We are given the equation of the curve:

`y = sqrt(8x - x^2)`

We want to solve for `(dy)/(dx)` . This could be done using the chain rule:

`(d(f(g(x))))/dx = f'(x)*g'(x)`

Hence,

`(dy)/(dx) = [1/2 (-x^2 + 8x)^(-1/2) ] * (-2x + 8)`

This simplifies to:

`(-x + 4)/(sqrt(x^2 + 8x))` [1]

dy/dx = (-x + 4)/(sqrt(x^2 + 8x))


For the stationary point, we know that `(dy)/(dx) = 0` . Since we already know the derivative of the equation (equation 1), we simply equate it to 0.

`(-x + 4)/(sqrt(x^2 + 8x)) = 0`

This implies that `-x + 4 = 0` and `x = 4` .

To get the value of y, we use the original equation of the curve, and plug in this x value:

`y = sqrt(8x - x^2) = sqrt(8*4 - 4^2) = sqrt(32 - 16) = sqrt(16) = 4`

Hence, the stationary point (x,y) for which the derivative is 0 is the point (4, 4).

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