What is the equation of the tangent line of #f(x) =(xlnx+4)/(xe^x-3)# at #x=6#?

Question

Emilia Aguilar · Accepted Answer

#y=((-22 e^6-(36 e^6+3) ln 6 -3)/(6 e^6 -3)^2)x #
#-6*((-22 e^6-(36 e^6+3) ln 6 -3)/(6 e^6 -3)^2)+(6 ln 6+4)/(6 e^6 -3)# Also equivalent to #y=0.005972344900598 x + 0.041935460457288#

Elijah Abram · Answer

undefined To find the equation of the tangent line of $ f(x) = \frac{x \ln x + 4}{x e^x - 3} $ at $ x = 6 $, first, we need to find the derivative of $ f(x) $. Then, we evaluate the derivative at $ x = 6 $ to find the slope of the tangent line. After that, we use the point-slope form of the equation of a line, $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point of tangency and $ m $ is the slope, to find the equation of the tangent line.

First, let's find the derivative of $ f(x) $:
$$ f'(x) = \frac{(x e^x - 3)(\frac{d}{dx}(x \ln x + 4)) - (x \ln x + 4)(\frac{d}{dx}(x e^x - 3))}{(x e^x - 3)^2} $$
$$ = \frac{(x e^x - 3)((\ln x + 1) + x \cdot \frac{1}{x}) - (x \ln x + 4)(e^x + x e^x)}{(x e^x - 3)^2} $$
$$ = \frac{(x e^x - 3)(\ln x + 1 + 1) - (x \ln x + 4)(e^x + x e^x)}{(x e^x - 3)^2} $$
$$ = \frac{(x e^x - 3)(\ln x + 2) - (x \ln x + 4)(e^x + x e^x)}{(x e^x - 3)^2} $$

Now, we evaluate $ f'(6) $ to find the slope of the tangent line at $ x = 6 $:
$$ f'(6) = \frac{(6 e^6 - 3)(\ln 6 + 2) - (6 \ln 6 + 4)(e^6 + 6 e^6)}{(6 e^6 - 3)^2} $$

After finding $ f'(6) $, we can then use the point-slope form of the equation of a line to find the equation of the tangent line.