Question

What is the equation of the line tangent to `f(x)=e^(3x-4)sine^x` at `x=lnpi`?

Answer 1

`y=-pi^4/e^4`

Explanation:

`f'(x)` is given by
`f'(x)=e^(4x-4)*cos(e^x)+3e^(3x-4)*sin(e^x)`
and we get
`f(ln(pi))=0` since

`sin(e^log(pi))=sin(pi)=0`
and
`f'(ln(pi))=-pi^4/e^4`