Here, if we think of f(x) as 5x^4 , then this expression gives the derivative of 5x^4 at the point x = . The derivative of 5x^4 is f′(x) = 20x^3.
At x = , we get f′ = 20= = .- A Use the distance formula (D^2 = (x − x 1 )^2 + (y − y 1 )^2 ) to determine the distance between the
curve and the point: D^2 = (x − 1)^2 + (y − 2)^2 = x^2 − 2x + 1 + y^2 − 4 y + 4. From the equation ofthe curve, you can solve for y^2 and y: y^2 = 9 − x^2 and y = . Substitute these backinto the formula for distance: D^2 = x^2 − 2x + 1 + 9 − x^2 − .To simplify calculations, recall that the functions D^2 and D will be minimized at the samepoint, so instead of solving for D, continue calculations with D^2 , which will now be called L:D^2 = L. Thus, L = 14 − 2x − . To minimize the distance, take the first derivative of Land set it equal to zero: = −2 + = 0. Solve this equation for x: x = ± . Findthe corresponding y-values: y = ± 2. The point closest to (1, 2) would then be . Youcan verify this with the second derivative test.- C If y = loga u then = . In this case, = = .
- B When dealing with parametric functions, your task is to eliminate t. In this case, notice y = 4t^9
= (2t^3 )^2 . Since x = 2t^3 , y = x^2.- B Use L’Hôpital’s Rule.