d/dx((x+2)2)
2(x+2) or 2x+4
d/dx(sin(x)/cos(x))
sec2(x)
d/dx(e4x)
4e4x
d/dx((x2+2x+1)21)
21(2x+2)(x2+2x+1)20
d/dx(2cos(x)+3sin(x))
-2sin(x)+3cos(x)
d/dx(x3e3x)
3x2e3x(x+1)
d/dx(x/x2)
-1/x2
d/dx(epi)
0
5x4/(x+1)6
d/dx(tan(2x))
2sec^2(x)
d/dx(sec2(x))
2sec2(x)tan(x)
d/dx(ex/(2x+3))
(ex(2x+3)-ex(2))/(2x+3)2
Using the limit definition of derivative, d/dx(3x-x2) looks like this (just set it up, do not compute)
limh->0 [(3(x+h)-(x+h)2)-(3x-x2)]/h
d/dx(e(x^3+3))
3x2e(x^3+3)
d/dx (cot(x))
-csc^2(x)
d/dx(sqrt(x)*12x)
18sqrt(x)
d/dx((x2+3x+1)/(x3−2x))
−(x4+6x3+5x2−2)/(x3−2x)2
If I asked you to determine the speed (in miles per hour) of a car whose position, where t=hours, is defined by
p(t)=t2+3t+2
what would its speed be exactly 5 hours after it started?
13 miles per hour
x(3x-4)/2(sqrt(x3-2x2+3))
d/dx(-ex(sin(x))
-excos(x)-exsin(x)
d/dx(sec(x)*tan(x))
sec(x)tan2(x)+sec3(x)
d/dx((x3+2x)/(x2−4))
(x4−14x2−8)/(x2−4)2
find the equation of the tan line of f(x) = x2 at the point (2,4)
y=4x−4
d/dx(tan(sqrt(x2+ex)))
-sin(sqrt(x2+ex)*(2x+ex)/2sqrt(x2+ex)
Find the equation of the tangent ine to g(x)=cos(3x) at the point (pi/6,0)
y=-3x+pi/2