1. find the derivative of (cosx)^(x^2).
2. find the x-coordinate of the point on the function: y=x^2 - 2x + 3 at which the tangent line is perpendicular to the line x+3y+3=0.
2. find the x-coordinate of the point on the function: y=x^2 - 2x + 3 at which the tangent line is perpendicular to the line x+3y+3=0.
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y=(cosx)^x^2 ==> lny = x^2*ln cosx and y´/y= 2x lncosx+x^2 (-tan x)
so y´= (cosx)^x^2[ 2x ln cox -x^2*tan x)
Call the poin (xo,yo) the tangentn at this point is y-(xo^2-2xo+3)= (2xo-2)(x-xo)
so as the slope of the line is -1/3 the slope of the tangent is 3= 2xo-2 so xo= 5/2
and yo= 25/4-5+3 = 17/4
so y´= (cosx)^x^2[ 2x ln cox -x^2*tan x)
Call the poin (xo,yo) the tangentn at this point is y-(xo^2-2xo+3)= (2xo-2)(x-xo)
so as the slope of the line is -1/3 the slope of the tangent is 3= 2xo-2 so xo= 5/2
and yo= 25/4-5+3 = 17/4