What is the equation of the tangent line of f(x) =sqrt(tanx-sinx)f(x)=√tanx−sinx at x = pi/4x=π4?

1 Answer
mason m
Sep 7, 2017

y-sqrt(1-1/sqrt2)=(2sqrt2-1)/(2sqrt(2-sqrt2))(x-pi/4)y−√1−1√2=2√2−12√2−√2(x−π4)

Explanation:

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

f(pi/4)=sqrt(tan(pi/4)-sin(pi/4))=sqrt(1-1/sqrt2)f(π4)=√tan(π4)−sin(π4)=√1−1√2

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))(π4,√1−1√2).

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/4x=π4. I'd start by writing the square root as a power of 1/212, and then differentiating the function using the chain rule.

f(x)=(tanx-sinx)^(1/2)f(x)=(tanx−sinx)12

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)

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