d(cosu)=
-sinu*du
y=sin(2x), dy/dx=
2cos (2x)
y=sec^2x, dy/dx=
2secxsecxtanx or2sec^2xtanx
Find the Equation of the Normal line at
x=pi/3
for f(x)=cos(x)
y=(2sqrt3*x)/3-(4pisqrt3-9)/18 or1.155x-0.709
d(cotu)=
-csc^2(u)*du
y=tan(3-x), dy/dx=
-sec^2(3-x)
y=cotxsecx, dy/dx=
cotxsecxtanx-secxcsc^2x or secx-secxcsc^2x
Find the Equation of the Tangent Line at
x=pi
for f(x)=tan(x)
y=x-3.142 or y=x-pi
d(cscu)=
-cscu*cotu*du
y=csc(x^2), dy/dx=
-2xcsc(x^2)cot(x^2)
y=sinx/cosx, dy/dx=
(cos^2x+sin^2x)/cos^2x or 1/cos^2x orsec^2x
If f(x)=sec(3x), find
f'(pi)
0
d(tanu)=
sec^2(u)*du
y=sec(pi+3x), dy/dx=
3sec(pi+3x)tan(pi+3x)
x=cscy, dy/dx=
-1/(cscycoty
If f(x)=csc(2x), find:
f'(pi/6)
-1.333 or
-4/3
d(secu)=
secu*tanu*du
y=cot(x/2), dy/dx=
-1/2csc^2(x/2)
Final Jeopardy. With your calculator, find the derivative of:
xy+secy=1/4
(-y(cos(y))^2)/(x(cos(y))^2+sin(y)
If
f(x)=sqrtsinx
Find
f'(pi/4)
0.420 or
2^(3/4)/4