How do you find dy/dxdydx by implicit differentiation of tan(x+y)=xtan(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)dydx=1sec2(x+y)sec2(x+y)

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

Explanation:

When doing implicit differentiation, you follow these essential steps:

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

So, let's differentiate both sides:

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

The right hand side just comes out as 11, 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]ddx[tan(x+y)]ddx[x+y]

This comes out to:

sec^2(x + y) * (1 + dy/dx)sec2(x+y)(1+dydx)

Putting this back in the whole equation:

sec^2(x + y) + sec^2(x + y)dy/dx = 1sec2(x+y)+sec2(x+y)dydx=1

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

dy/dx = [1-sec^2(x + y)]/sec^2(x + y)dydx=1sec2(x+y)sec2(x+y)

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

dy/dx = [1-sec^2(0)]/sec^2(0)dydx=1sec2(0)sec2(0)

sec(0)sec(0) is simply 1/cos(0)1cos(0). Since cos(0)cos(0) = 1, sec(0)sec(0) is also 1. So, wherever we see sec(0)sec(0), we just plug in 11:

dy/dx = [1-1]/1 = 0dydx=111=0

So your tangent line would have a slope of 00 at the point (0,0)(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)-1dydx=cos2(x+y)1
Evaluating at (0,0)(0,0) gives dy/dx=0dydx=0

Explanation:

Implicit differentiation is just differentiation using chain rule.

(d tan(x+y))/(d(x+y)) xx (d (x+y))/dxdtan(x+y)d(x+y)×d(x+y)dx=dx/dxdxdx
sec^2(x+y)xx(1+dy/dx) = 1sec2(x+y)×(1+dydx)=1

Rearranging:

1+dy/dx = cos^2(x+y)1+dydx=cos2(x+y)
dy/dx = cos^2(x+y)-1 dydx=cos2(x+y)1

Substituting (0,0)(0,0) in above equation,
we get:
dy/dx = cos^2(0)-1dydx=cos2(0)1
i.e dy/dx=0dydx=0