How do you find #dy/dx# by implicit differentiation of #tan(x+y)=x# and evaluate at point (0,0)?

2 Answers
Jan 26, 2018

#dy/dx = [1-sec^2(x + y)]/sec^2(x + y)#

At #(0,0)#, #dy/dx = 0#

Explanation:

When doing implicit differentiation, you follow these essential steps:

  1. Take the derivative of both sides of the equation with respect to #x#.
  2. Differentiate terms with #x# as normal.
  3. Differentiate terms with #y# as normal too but tag on a #dy/dx# to the end.
  4. Solve for the #dy/dx#.

So, let's differentiate both sides:

#d/dx[tan(x + y)] = d/dx[x]#

The right hand side just comes out as #1#, but the left hand side will require that we use a chain rule. This goes as:

#d/dx[tan(x + y)] * d/dx[x + y]#

This comes out to:

#sec^2(x + y) * (1 + dy/dx)#

Putting this back in the whole equation:

#sec^2(x + y) + sec^2(x + y)dy/dx = 1#

Now, you just solve for #dy/dx# using some basic algebra:

#dy/dx = [1-sec^2(x + y)]/sec^2(x + y)#

Now, you're given the point #(0,0)# to evaluate the derivative at. All you do is plug this in:

#dy/dx = [1-sec^2(0)]/sec^2(0)#

#sec(0)# is simply #1/cos(0)#. Since #cos(0)# = 1, #sec(0)# is also 1. So, wherever we see #sec(0)#, we just plug in #1#:

#dy/dx = [1-1]/1 = 0#

So your tangent line would have a slope of #0# at the point #(0,0)#.

If you want more help in implicit differentiation, check out my video:

Hope that helped :)

Jan 26, 2018

#dy/dx = cos^2(x+y)-1#
Evaluating at #(0,0)# gives #dy/dx=0#

Explanation:

Implicit differentiation is just differentiation using chain rule.

#(d tan(x+y))/(d(x+y)) xx (d (x+y))/dx#=#dx/dx#
#sec^2(x+y)xx(1+dy/dx) = 1#

Rearranging:

#1+dy/dx = cos^2(x+y)#
#dy/dx = cos^2(x+y)-1 #

Substituting #(0,0)# in above equation,
we get:
#dy/dx = cos^2(0)-1#
i.e #dy/dx=0#