Question

What is the equation of the tangent to the curve `y=9tanx` at the point where `x=(2pi)/3`?

Answer 1

`y = 36x-24pi -9sqrt(3)`

Explanation:

`y=9tanx`

Differentiating wrt `x` gives:
`dy/dx =9sec^2x`

So, When `x=(2pi)/3 => y=9tan((2pi)/3) =-9sqrt(3)`
and, `dy/dx = 9sec^2((2pi)/3)=9(-2)^2=36`

Thus, the tangent has gradient `m=36` and passes through `((2pi)/3,-9sqrt(3))`

So, using `y-y_1=m(x-x_1)` the required equation is:
`y-(-9sqrt(3))) = 36(x-(2pi)/3)`
`:. y+9sqrt(3) = 36x-(72pi)/3`
`:. y = 36x-24pi -9sqrt(3)`