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

1 Answer
Sep 7, 2017

#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)#