Chain Rule
Differentiate
y=sqrt(2x)
1/(sqrt(2x)
Differentiate
x=y^5+y
dy/dx=1/(5y^4+1)
Differentiate
y=arcsin(2x)
y'=2/(sqrt(1-4x^2)
Differentiate
y=x5^x
y'=5^x+ln(5)x5^x
Find the slope of the tangent line of f(x) when x=0.
f(x)=(2x-1)^3
f'(x)=6(2x-1)^2
f'(0)=6
Differentiate
y=sin^2x
y'=2sinxcosx
Differentiate
xy=y
dy/dx=y/(1-x)
Differentiate
f(x)=arccos(2x-1)
f'(x)=-2/(sqrt(1-(2x-1)^2
Differentiate
f(x)=6^(x^2)
f'(x)=2ln(6)x6^(x^2)
Find the slope of the tangent line of f(x) when x=e.
f(x)=x^x
f'(x)=x^x(lnx+1)
f'(e)=2e^e
Differentiate
y=(x^2+4)^(3/2)
y'=3xsqrt(x^2+4)
Find y'.
xe^y=4x^2
y'=(8x-e^y)/(xe^y)
Differentiate.
g(r)=4tan^-1(r/2)+r
g'(r)=8/(r^2+4)+1
Differentiate
g(t)=t^2-log2t
g'(t)=2t-1/(ln10t)
Consider the curve below. Find all the points where the tangent line to the curve has a slope of 1/2.
y^2 = 2+xy
(0,sqrt2)
(0,-sqrt2)
Differentiate
y=ln(tan^2x)
2sec^2x/(tanx)
Differentiate
tan(xy)=x
dy/dx=(1-ysec^2(xy))/(xsec^2(xy))
Differentiate.
y=sqrt(1-x^2)(sin^-1x)
y'=1-(xsin^-1x)/(sqrt(1-x^2)
Differentiate
f(x)=log_2(sin(2x))
f'(x)=2cot(2x)/(ln2)
Consider the function f=y y>0 whose curve is given by the equation below. Write an equation for the line tangent to the curve at the point:
(0,sqrt3)
2y^2-6=ysinx
y=sqrt3+1/4x
Differentiate
f(x)=e^(x^2cosx)
f'(x)=-xe^(x^2cosx)(xsinx-2cosx)
Find y''.
y^3+y=x^2
y''=(2(3y^2+1)^2-24x^2y)/(3y^2+1)^3
Differentiate.
cot^-1(xy)=y
dy/dx=y/(-1-x-x^2y^2)
Differentiate.
log_3(x+y)=8x
dy/dx=8ln3(x+y)-1
Consider the curve below. Write an equation for each horizontal tangent line the curve has. (You can use a graphing calculator to solve for y)
2y^3+6x^2y-12x^2+6y=1
y=0.165