Question

What is the equation of the tangent line of `f(x) =((x-1)^3(x^2+2))/(e^x-1)^4` at `x=2`?

Answer 1

`y = f'(2)x + (f(2) - 2f'(2))`
with `f(x)` as shown in the question stem and `f'(x)` as derived below

Explanation:

The instantaneous slope of the tangent to the function will be its first derivative. Differentiation is a science, integration is an art. That is a messy derivative.

Gritting teeth!

The numerator is the product of two functions, one of them composite.

The first of the two functions in the numerator has derivative
`3 (x-1)^2 (1) = 3(x - 1)^2` (chain rule)

The second of the two functions in the numerator has derivative
`2x`

So, the overall derivative of the numerator is
`3(x-1)^2 (x^2 + 2) + 2 x (x - 1)^3` (product rule)

The derivative of the denominator is
`4e^x(e^x - 1)^3` (chain rule)

So, the derivative of the whole function is

`f'(x) = ((3(x-1)^2 (x^2 + 2) + 2 x (x - 1)^3) (e^x - 1)^4 - (x - 1)^3(x^2 + 2) 4e^x(e^x - 1)^3)/((e^x - 1)^4 )^2`

`= ((e^x - 1)^3(x-1)^2((3 (x^2 + 2) + 2 x (x - 1)) (e^x - 1) - 4 (x - 1)(x^2 + 2) e^x))/(e^x - 1)^8`

`= ((x-1)^2((3 (x^2 + 2) + 2 x (x - 1)) (e^x - 1) - 4(x - 1)(x^2 + 2) e^x))/(e^x - 1)^5`

`= ((x-1)^2((3 x^2 + 6 + 2 x^2 - 2x) (e^x - 1) - 4(x^3 - x^2 + 2x - 2) e^x))/(e^x - 1)^5`

`= ((x-1)^2((5 x^2 - 2 x+ 6)e^x - (3 x^2 + 6 + 2 x^2 - 2x) - (4x^3 - 4x^2 + 8x - 8)e^x))/(e^x - 1)^5`

`= - ((x-1)^2((4x^3 -9 x^2 + 10 x -14)e^x + 5 x^2 - 2x + 6 ))/(e^x - 1)^5`

(quotient rule)

I used this site to check workings for the derivative (I think it only sensible!)

It is noted that a graph of the function will pass through the point `x = 2`, `y = f(2)` and that its instantaneous slope at this point will be `f'(2)`

so, using the formula for a line passing through some point `(x_1, y_1)` of slope `m`: `(y - y_1) = m (x - x_1)`, the tangent line of `f(x)` at `x = 2` will satisfy

`(y - f(2)) = f'(2)(x - x_1)`

that is

`y = f'(2)x + (f(2) - 2f'(2))`

I will not evaluate f(2) or f'(2)!

Answer 2

The equation is nearly `y = -0.00189x + 0.00738`

Explanation:

Alternatively we may use logarithmic differentiation to simplify the process a little.

We start by seeing that

`f(2) = ((2 - 1)^3(2^2 + 2))/(e^2 - 1)^4 = 6/(e^2 - 1)^4`

Now we differentiate

`lny = ln(((x -1)^3(x^2 +2))/(e^x - 1)^4)`

`lny = ln(x - 1)^3 + ln(x^2 + 2) - ln(e^x - 1)^4`

`lny = 3ln(x- 1) + ln(x^2 + 2)- 4ln(e^x -1)`

`1/y(dy/dx) = 3/(x -1) + (2x)/(x^2 + 2) - (4e^x)/(e^x - 1)`

`dy/dx = (((x - 1)^3(x^2 + 2)^3)/(e^x- 1)^4)(3/(x -1) + (2x)/(x^2 + 2) - (4e^x)/(e^x -1))`

So the slope of the tangent at `x = 2` is

`m = 6/(e^2 - 1)^4 * (3 + 2/3 - ((4e^3)/(e^2 - 1))`

`m = 6/(e^2 - 1)^4 * (11/3 - (4e^3)/(e^2 - 1))`

`m = (6(11e^2 -11 - 4e^3))/(3(e^2 - 1)^5`

`m = (2(11e^2 - 11 - 4e^3))/(e^2 - 1)^5`

The equation of the tangent is therefore

`y - 6/(e^2 - 1)^4 = m(x - 2)`

Approximating we get about

`y = -0.00189(x - 2) + 0.0036`

`y = -0.00189x +0.00378 + 0.0036`

`y = -0.00189x + 0.00738`

Hopefully this helps!