Question

What is the equation of the line normal to `f(x)=ln(xe^(3x))` at `x=1`?

Answer 1

`y=(-1/4)x+13/4`

Explanation:

`f(x) = ln(xe^(3x))`
`f(1)= ln(1*e^3)=3ln(e)=3`
`f(x)=ln(x)+3x*ln(e)=ln(x)+3x`
`f'(x)=1/x+3`
`f'(1)=4` => slope of the tangent at x = 1:
Slope of the normal at x = 1 is: -1/4:
`m=-1/4, (x_1,y_1)=(1,3)`
`y-y_1=m(x-x_1)`
`y-3=-1/4(x-1)`
`y=(-1/4)x+13/4`

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