Question

What is the equation of the line normal to `f(x)=ln(x+xe^(3x))` at `x=2`?

Answer 1

`y - ln(2+2e^6) = -(2+2e^6)/(1+7e^6) (x - 2)`

Explanation:

We can rewrite the equation as follows (and consequently avoid needing to use the product rule):

`f(x) = ln(x+xe^(3x))`
`:. f(x) = ln(x(1+e^(3x)))`
`:. f(x) = ln(x) + ln(1+e^(3x))`

We can differentiate (using the chain rule):
`:. f'(x) = 1/x + 1/(1+e^(3x))*(3e^(3x))`
`:. f'(x) = 1/x + (3e^(3x))/(1+e^(3x))`
`:. f'(x) = ( (1+e^(3x)) + (3xe^(3x)) ) / (x(1+e^(3x)))`

`:. f'(x) = ( 3xe^(3x) + e^(3x) + 1) / (x + xe^(3x))`

When `x=2=>f(2)=ln(2+2e^6)`
And, `f'(2)=(6e^6 + e^6 +1)/(2+2e^6) = (1+7e^6)/(2+2e^6)`

This is the gradient of the tangent when `x=2`. As the normal and tangent are perpendicular the product of their gradients is `-1`, so the normal passes through `(2,ln(2+2e^6))` and it has gradient `m=-(2+2e^6)/(1+7e^6)`

So using `y-y_1=m(x-x_1)` the required equation is;

`y - ln(2+2e^6) = -(2+2e^6)/(1+7e^6) (x - 2)`