Question

Determine the equation of the tangent to y = (3x^-2 - 2x^3)^5 at (1 , 1 )?

Answer 1

The equation of the tangent is `y = -60x + 61`.

Explanation:

You were right to assume you should use the chain rule. The first step is always to reverse any compositions that could have taken place to give you the function that you must differentiate.

Let `y = u^5`, then `u = 3x^-2 - 2x^3`. We have our two functions! Now, we differentiate.

`y' = 5u^4` and `u' = -6x^-3 - 6x^2`

The chain rule states that

`dy/dx = dy/(du) * (du)/dx`

`dy/dx = 5u^4 * -6x^-3 - 6x^2`

`dy/dx = 5(3x^-2 - 2x^3)(-6x^-3 - 6x^2)`

This will not have to be simplified, since our goal here isn't to find the derivative, but instead the equation of the tangent. Think of it as a stopover on a flight.

The slope of the tangent is obtained by evaluating the point `x = a` into the derivative.

`(dy/dx)_(x = 1) = 5(3(1)^-2 - 2(1)^3)(-6(1)^-3 - 6(1)^2)`

The nice thing about `1` is that `1^n = 1` for all values of `n`.

`(dy/dx)_(x=1) = 5(1)(-12)`

`(dy/dx)_(x=1) = -60`

The equation of the tangent is given by

`y - y_1 = m(x - x_1)`

`y - 1 = -60(x - 1)`

`y - 1= -60x + 60`

`y = -60x + 61`

Hopefully this helps!