What is the equation of the tangent line of #f(x) = xtanx-(xcosx)/(1-x)# at # x = pi/4#?

This problem has three major steps to it:

1. Use the Product & Quotient rules to calculate the derivative.

2. Use the derivative to find the slope at #x=pi/4#, and use the original function to find #f(pi/4)#

3. Put them all together to write the equation of the line.

First thing we need to do is utilize the quotient and product rules to find the derivative of the function. This is not as difficult as it seems, but is still tedious to format, so I have simply calculated it on paper, and attached a picture of that here:

Now that we have the derivative, we can calculate the slope at any point of the original function using it. So, let's plug in #pi/4#:

#f'(pi/4) = tan(pi/4) + xsec^2(pi/4) - (cos(pi/4)-xsin(pi/4)+xcos(pi/4))/((1-pi/4)^2)#

#= 1 + pi/4(2/sqrt(2))^2 - (sqrt(2)/2-pi/4(sqrt(2)/2)+pi/4(sqrt(2)/2))/((1-pi/4)^2)#

= #1 + pi/2 - (sqrt(2)/2)/(1-pi/4)^2#

You could simplify that further if you wished, but for now this is fine.

Now, we'll need to calculate #f(pi/4)#:

#f(pi/4) = (pi/4)(1) - (pi/4(sqrt(2)/2))/(1-pi/4)#

= #(pi/4) - (pisqrt(2)/8)/(1-pi/4)#

Now that we have that, all we need to do is plug in everything into point slope form and we have:

#y = (1 + pi/2 - (sqrt(2)/2)/(1-pi/4)^2)(x-pi/4) + ((pi/4) - (pisqrt(2)/8)/(1-)pi/4)#

I am so sorry this answer is so full of equations....but it had to be done. Nonetheless, hope that helped :)