Question

How do you find the equation of tangent line to the curve `f(x) = 9tanx` at the point `(pi/4, f(pi/4))`?

Answer 1

`36x-2y+18-9pi=0`

Explanation:

We have to find equation of tangent to curve `f(x)=9tanx` at point `(pi/4,f(pi/4))` i.e. `(pi/4,9tan(pi/4))` and as `tan(pi/4)=1`, this point is `(pi/4,9)`.

The slope of tangent is given by value of first derivative of `f(x)`, at that point I.e. at `(pi/4,9)`.

As `(df)/(dx)=9sec^2x` we have `f'(pi/4)=9sec^2(pi/4)=9xx(sqrt2)^2=18`

As slope is `18` and tangent passes through `(pi/4,9)`, its equation is

`(y-9)=18(x-pi/4)` i.e. `36x-2y+18-9pi=0`.

graph{(36x-2y+18-9pi)(y-9tanx)=0 [-4, 4, -20, 20]}