Question

What is the equation of the tangent line of `f(x) =ln(sinx+3x)` at `x=pi/4`?

Answer 1

`1.2102 x-y+0.1707=0`

Explanation:

y=f(x)=ln(sinx+3x)=ln(1/sqrt2+3/4pi)=1.1195, nearly, at `x=pi/4`.

`y'=(cosx+3)/(sinx+3x)=(1/sqrt2+3)/(1/sqrt2+3/4pi)`

`=1.2102, at x = pi/4`.

So, the equation to the tangent atP(1.1195, 0.7854) is

`y-1.1195=1.2102(x-0.7854)`, giving

`1.2102x-y+0.1707`, nearly.