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Implicit Differentiation
Chain Rule
Inverse Derivatives
Expo/Log Deriviatives
Related Rates
100
x(^3)+x(^2)y+9y^(2)=6
-x(3x+2y)/(x^(2)+8y)=y'
100
y=e^(√x)
y'=e^(√x)/(2√x)
100
y=sin^(-1)(x^(2))
y'=(2x)/√(1-x^(4))
100
f(x)=ln(cos(x))
f'(x)=-tan(x)
100
If V is the volume of a cube with edge length x and the cube expands as time passes, find (dv/dt) in tems of (dx/dt).
dv/dt= 3x^(2) dx/dt
200
x^(2)y+xy^(2)=3x
y'=(3-2xy-y^(2))/(x^(2)+2xy)
200
f(x)=(4√)(1+2x+x^(3))
f'(x)=(2+3x^(2))/[4(1+2x+x^(3)^(3/4)]
200
H(x)=(1+x^(2))arctan(x)
H'(x)=1+2xarctan(x)
200
y=ln(x)/(1+x)
y'=(1+x-xln(x))/[x(1+x)^(2)]
200
If y=x^(3)+2x and dx/dt=5, find dy/dt when x=2.
70
300
√(xy)=1+x^(2)y
y'=(4xy√(xy)-y)/(x-2x^(2)√(xy))
300
DAILY DOUBLE g(t)=1/(t^(4)+1)^3
g'(t)=-12t^(3)/(t^(4)+1)^4
300
y=arcsin(tan(x))
y'=sec^(2)x/√1-tan^(2)x
300
y=ln(e^(-x)+xe^(-x)
y'=-x/(1+x)
300
Two cars start moving from the same point. One travels south at 60mph and the other travels west at 25mph. At what rate is the distance between the cars increasing two hours later?
65mph
400
4cos(x)sin(y)=1
y'=tan(x)tan(y)
400
y=e^(xcos(x))
y'=(cos(x)-xsin(x))e^(xcos(x))
400
y=2√(x)arctan√(x)
y'=1/(1+x)+(arctan(√(x))/√(x)
400
DAILY DOUBLE y=x^(sin(x))
y'=x^(sin(x))[sin(x)/x+ln(x)cos(x)]
400
Two sides of a triangle are 4m and 5m in lengthe and the angle between them is increasing at a rate of .06 radians/second. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3.
3 m^(2)/seconds
500
cos(x-y)=xe^(x)
y'=1+e^(x)(1+x)/sin(x-y)
500
y=cot^(2)(sin(x))
-2cos(x)cot(sin(x))csc^(2)(sin(x))
500
f(x)=e^(x)-x^(2)arctan(x)
f'(x)=e^(x)-x^(2)/(1+x^(2))-2xarctan(x)
500
f(x)=x/(1-ln(x+1))
f'(x)=[2x-1-(x-1)ln(x-1)]/(x-1)[1-ln(x-1)]^(2)
500
A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance of 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m?
7√(15)/4 which is about 6.78 m/s