Question

Question #15785

Answer 1

`-12/5`

Explanation:

To make the function easier to derive we will isolate `y`.
`5lny=12-3x^2`
`lny={12-3x^2}/5`
`y=e^{{12-3x^2}/5}`

The we find the tangent line, which will be the derivative
`y'=e^{{12-3x^2}/5}(-{6x}/5)`

We then plug in `x=2` to get
`y'(2)=e^{{12-3(2)^2}/5}(-{6(2)}/5)`
`y'(2)=e^{0/5}(-12/5)`
`y'(2)=1(-12/5)`
`y'(2)=-12/5`