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

Also equivalent to

By signing up, you agree to our Terms of Service and Privacy Policy

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.

By signing up, you agree to our Terms of Service and Privacy Policy

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7