y' = (sec(x))^2
x = yy differentiate both sides with respect to x dx = (y * yy-1) dy dy/dx = y * yy-1 dy/dx = yy = x hence differentiate of y wrt x is x only
8
You have to know what x and y equals first!
if the x equals -1.2 and the y is -4, 4 what is the slope
y' = (sec(x))^2
This is definitely false; if x=2 and y=3, x to the y power is 8, but y to the x power is 9, which are not equal.
999 and 1
x = yy differentiate both sides with respect to x dx = (y * yy-1) dy dy/dx = y * yy-1 dy/dx = yy = x hence differentiate of y wrt x is x only
8
if y = 2x then x = log2 y
You have to know what x and y equals first!
if the x equals -1.2 and the y is -4, 4 what is the slope
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
Yes, if x10=y.
Well you can find what it equals in terms of x. If you know that y plus y equals 25 to the power of 25 then you know that 2y=25^25 which means that y=(25^25)/2 - this is a huge number. To find x+y you just add x to what y equals so you get x+(25^25)/2
It is y = 0