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

##### 1 Answer

#### Explanation:

Find the

#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

We also need to find the slope of the tangent line, which we can do by evaluating the derivative of the function at

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

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

Note that

#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

#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

#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

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

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