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How do you prove that the derivative of sec x is equal to sec x tan x?
Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.
Formula to calculate v-belt length?
The answer is: L = pi x (D + d)/2 + 2 x ( C x Cos(a) + a x (D-d)/2) where a = arcsin(D-d)/(2 x C) in radians. Where C is the center distance, D is the large pulley diameter, and d is the small pulley diameter.
What does d mean in 1 x c x 3 equals d?
d=3c
How to differentiate x plus 2?
d/dx(x + 2) = d/dx(x) + d/dx(2) = 1 + 0 = 1
What is the second derivative of the natural logarithm of 1 minus X?
d/dx ln(1-x) = 1/(1-x) d/dx 1/(1-x) = 1-x * d/dx 1 - 1 * d/dx(1-x)/(1-x^2 = 1-x * 0 - 1 * 0 - 1/(1-x)^2 = -1/(1-x)^2 --------------
