Hi,
Q1. find equation of the tangent line and the normal line to the curve
y=(3x-2)e^-x at the point (0,-2).
Q2. Find derivative of cos(x+y) + y = 0
thanks,
moon
Hi,
Q1. find equation of the tangent line and the normal line to the curve
y=(3x-2)e^-x at the point (0,-2).
Q2. Find derivative of cos(x+y) + y = 0
thanks,
moon
to Q1:
1, Use product and chain rule to calculate the first derivative ($\displaystyle y'=(5-3x)e^{-x}$)
2. Plug in x = 0 to get the slope of the tangent (m = 5)
3. Use point-slope-formula of the straight line to get the equation of the tangent (y+2=5(x-0) )
4. Calculate the negative reciprocal of m to get the slope of the normal to the curve in (0, -2): ($\displaystyle m_\perp = -\dfrac15 $)
5. Use point-slope-formula to get the equation of the normal: ($\displaystyle y+2 = -\dfrac15 (x-0)$ )
6. Now it's your turn!
to Q2:
This is an implicite function:
$\displaystyle f(x,y) = \cos(x+y)+y=0$
You are asked to calculate the drivative:
$\displaystyle \dfrac{dy}{dx}=y'= - \dfrac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$
$\displaystyle \dfrac{\partial f}{\partial x} = -\sin(x+y)$
$\displaystyle \dfrac{\partial f}{\partial y} = 1-\sin(x+y)$
Therefore:
$\displaystyle \dfrac{dy}{dx}=y'= \dfrac{\sin(x+y)}{1-\sin(x+y)}$