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

Question

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

Scarlett Allison · Accepted Answer

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

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) #

Emma Aldridge · Answer

undefined To find the equation of the line normal to the function f(x) = ln(x + xe^(3x)) at x = 2, we need to determine the slope of the tangent line at x = 2 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can use the derivative of the function. Taking the derivative of f(x) with respect to x, we get:

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

Evaluating f'(2), we find:

f'(2) = (1 + e^(3*2))/(2 + 2e^(3*2))

Simplifying this expression, we have:

f'(2) = (1 + e^6)/(2 + 2e^6)

Now, to find the slope of the normal line, we take the negative reciprocal of f'(2):

slope of normal line = -1/f'(2)

Finally, we have the equation of the line normal to f(x) at x = 2:

y - f(2) = slope of normal line * (x - 2)