Derivatives
g(x)=7/2x^2-3x+12
g'(x)=7/2x-3
y=(4x^2+3)(2x+5)
y'=2(4x^{2} + 3) + 8x(2x + 5)
x^2-4xy+y^2=4
y'=\frac{2x-4y}{4x+2y}
The area of a triangle with side lengths a and b and angle theta is given by A=1/2ab sin \theta .
If a=2 cm, b=3 cm, and theta increases at a rate of 0.2 rad/min, How fast is the area increasing when \theta = \pi/3 ?
\frac{dA}{dt}=0.3 cm^2/min
s(t)=1/t+1/t^2
s'(t)=-1/t^2-2/t^3
y=sin(cos(x))
y'=-cos(cos(x))sin(x)
\sqrt{x+y}=x^4+y^4
y'=\frac{8x^3-(x+y)^{-1/2}}{(x+y)^{-1/2}-8y^3}
Linear approx of f(x)=cos(2x) at a=\pi/6
L(x) = \frac{1}{2} - \sqrt{3}(x - \frac{\pi}{6})
g(x)=(x+2\sqrtx)e^x
g'(x)=\left(x + 2 \sqrt{x}\right) \mathrm{e}^{x} + \left(\frac{1}{\sqrt{x}} + 1\right) \mathrm{e}^{x}
f(z)=e^{z/(z-1)}
f'(z)=(\frac{1}{z - 1} - \frac{z}{(z - 1)^{2}})e^{\frac{z}{z - 1}}
Find the tangent line to 2(x^2+y^2)^2=25(x^2-y^2) at (3,1) .
y=-9/13x+40/13
Use Linear approx to approx 1/4.002
\frac{1}{4.002} \approx 0.249875