# Math Help - Derivatives and tangents

1. ## Derivatives and tangents

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

2. What experience of differentiation have you got?

You need some experience with this to solve these types of questions.

3. Originally Posted by Moon Hoplite
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 ( $y'=(5-3x)e^{-x}$)

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): ( $m_\perp = -\dfrac15$)

5. Use point-slope-formula to get the equation of the normal: ( $y+2 = -\dfrac15 (x-0)$ )

4. hi thanks, i know how to find derivatives, but i dont know how to expand
"cos(x+y) + y = 0" in terms of y, could anyone help on that question?

5. Originally Posted by Moon Hoplite
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 Q2:

This is an implicite function:

$f(x,y) = \cos(x+y)+y=0$

You are asked to calculate the drivative:

$\dfrac{dy}{dx}=y'= - \dfrac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

$\dfrac{\partial f}{\partial x} = -\sin(x+y)$

$\dfrac{\partial f}{\partial y} = 1-\sin(x+y)$

Therefore:

$\dfrac{dy}{dx}=y'= \dfrac{\sin(x+y)}{1-\sin(x+y)}$