differential

[MathPrince]

(1) y' = (1/2) sec^(2)[x/2]
y=8-x (so) m = 1 (so) 1 = (1/2) sec^(2)[x/2] (so) 2 = sec^(2)[x/2] (so) sec [x/2] = sq. r. (2) (so) x/2 = 45 (so) x = 90 / 270

(2) y = sq. r. [25 - x^(2)] (so) y' = [-2x] / 2 sq. r. [25 - x^(2)] (so) y' (at x=3) = -6 / 8 (so) m = -2/3
eq. of tang. [y - 4] / [x - 3] = -2/3 (so) 3y + 2x - 6 = 0
at x=0 "intersection with y-axis" y = 2 point is (0,2)
at y=0 "intersection with x-axis" x = 3 point is (3,0)
the area of the triangle = (1/2) (2) (3) = 3 unit square

(3) y = - [tan x] / [1- tan^(2) x] at x= 15 (so) y = - tan 2x = - tan 30 = - 1 / sq. r. (3)

(5) v(h) = [(5+h)^2 - 2(5+h) +3] - [25 - 10 + 3] = h^(2)
A(h) = v(h) / h = h
you can get the averae as an exercise

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