Question

What is the equation of the tangent line of `f(x) =sqrt(tanx-sinx)` at `x = pi/4`?

Answer 1

`y-sqrt(1-1/sqrt2)=(2sqrt2-1)/(2sqrt(2-sqrt2))(x-pi/4)`

Explanation:

Find the `y`-coordinate of the point the tangent line will pass through:

`f(pi/4)=sqrt(tan(pi/4)-sin(pi/4))=sqrt(1-1/sqrt2)`

There are a lot of ways to write this, so we'll just leave it alone. The tangent line passes through the point `(pi/4,sqrt(1-1/sqrt2))`.

We also need to find the slope of the tangent line, which we can do by evaluating the derivative of the function at `x=pi/4`. I'd start by writing the square root as a power of `1/2`, and then differentiating the function using the chain rule.

`f(x)=(tanx-sinx)^(1/2)`

`f'(x)=1/2(tanx-sinx)^(-1/2)d/dx(tanx-sinx)`

Note that `d/dxtanx=sec^2x` and `d/dxsinx=cosx`:

`f'(x)=1/2(tanx-sinx)^(-1/2)(sec^2x-cosx)`

`f'(x)=(sec^2x-cosx)/(2sqrt(tanx-sinx))`

So the slope of the tangent line at `x=pi/4` is:

`f'(pi/4)=(sec^2(pi/4)-cos(pi/4))/(2sqrt(tan(pi/4)-sin(pi/4))`

`f'(pi/4)=((sqrt2)^2-1/sqrt2)/(2sqrt(1-1/sqrt2))`

Multiplying through by `sqrt2/sqrt2`:

`f'(pi/4)=(2sqrt2-1)/(2sqrt2sqrt(1-1/sqrt2))`

`f'(pi/4)=(2sqrt2-1)/(2sqrt(2(1-1/sqrt2)))`

`f'(pi/4)=(2sqrt2-1)/(2sqrt(2-sqrt2))`

Now that we know the slope of the tangent line and a point it passes through, which is `(pi/4,sqrt(1-1/sqrt2))`, we can write the equation of the tangent line in point-slope form:

`y-y_1=m(x-x_1)`

`y-sqrt(1-1/sqrt2)=(2sqrt2-1)/(2sqrt(2-sqrt2))(x-pi/4)`