Derivatives and tangents

**Moon Hoplite** · January 6th 2009, 07:16 AM

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

**craig** · January 6th 2009, 07:32 AM

What experience of differentiation have you got?

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

**earboth** · January 6th 2009, 07:44 AM

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: ( $y+2 = -\dfrac15 (x-0)$ )

6. Now it's your turn!

**Moon Hoplite** · January 6th 2009, 09:51 PM

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?

**earboth** · January 6th 2009, 10:21 PM

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:

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

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