How do you find at equation of the tangent line to the graph #y=(cosx)/(sinx+2)# at x =pi/2?

1 Answer
Jan 19, 2017

#y = pi/6-1/3x#

Explanation:

Start by finding the y-coordinate of tangency.

#y = cos(pi/2)/(sin(pi/2) + 2)#

#y= 0/(1 + 2)#

#y = 0#

Now differentiate using the quotient rule.

#y' = (-sinx(sinx + 2) - cosx(sinx))/(2 + sinx)^2#

#y' = (-sin^2x - 2sinx - cosxsinx)/(2 + sinx)^2#

The slope of the tangent is given by evaluating your point #x= a# into the derivative.

#m_"tangent" = (-sin^2(pi/2) - 2sin(pi/2) - cos(pi/2)sin(pi/2))/(2+ sin(pi/2))^2#

#m_"tangent" = (-(1)^2 - 2(1) - 0(1))/(2 + 1)^2#

#m_"tangent" = (-1 - 2)/(2 + 1)^2#

#m_"tangent"= -3/9=-1/3#

We now find the equation of the tangent.

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

#y - 0 = -1/3(x - pi/2)#

#y = pi/6-1/3x #

Hopefully, this helps!