Question

What is the equation of the tangent line and normal line to y = 2x`e^x` at (0, 0)?

Answer 1

Tangent line equation: `y = 2x`
Normal line equation: `y = -1/2x`.

Explanation:

The derivative is given by the product rule

`y' = 2e^x + 2xe^x`

The slope of the tangent is therefore

`y'(0) = 2e^0 + 2(0)e^0 = 2`

Therefore the equation of the tangent line is

`y - y_1 = m(x - x_1)`

`y - 0 = 2(x - 0)`

`y = 2x`

The normal line is perpendicular to the tangent, so has slope `-1/2`.

`y - 0 = -1/2(x - 0)`

`y = -1/2x`

Here is a graphical representation. In red is the curve `y = 2xe^x`, in blue is the tangent line and in green is the normal line.

Hopefully this helps!

Answer 2

Tangent line: y = 2x
Normal line: y = `-1/2`x

Explanation:

STEP 1: Find the derivative of the given function y.
Use the Product Rule to differentiate the function.
y' = 2`e^x` + 2x`e^x`

STEP 2: Since we are looking for the equation of the tangent line and normal line at x=0, we want to find the derivative at that point. So find `y'(0)`:
y'(0) = 2`e^0` + 2*0`e^0` = 2+0 = 2

STEP 3: Write out the equation of the tangent line by using the slope found in step 2 and the point given.
`y - 0 = 2(x - 0)`
`y = 2x`

STEP 4: Knowing that the slope of the normal line is the negative reciprocal of the slope of the tangent line, we can find the slope of the normal line at the point (0,0).
`m_{NL} = -1/(m_{TL}) = -1/2`

STEP 5: Write out the equation of the normal line by using the slope found in step 4 and the point given.
`y - 0 = -1/2(x - 0)`
`y = -1/2 x`