Answer to Question #86314 in Calculus for Ridhi

Question

Write the equation of the line tangent to the curve sin(x+y) = x exp (x+y) at the origin (0,0).

Answer 1

Since we already know one of the points of the line (namely the origin, (0; 0)), finding the slope $k$ of the tangent line will be enough to write equation of this line as $y = kx.$

Slope of the tangent line can be found as the derivative of y with respect to x, i.e. $\frac{dy}{dx} .$

With the goal of finding this derivative, let's differentiate both sides of the curve equation with respect to x:

$\sin(x+y) = x\exp(x+y);$

$(1 + \frac{dy}{dx})\cos(x+y) = \exp(x+y) + x(1 + \frac{dy}{dx})\exp(x+y);$

We now solve this for $\frac{dy}{dx}$: $\frac{dy}{dx} = -1 + \frac{\exp(x+y)}{-x\exp(x+y)+\cos(x+y)} .$

This is the representation of the derivative expressed through both x and y, but this is fine for our purpose since we know both x and y of the point of interest: $x = y = 0.$

Entering these values of x and y, we get the value of the derivative at the point of origin $x = y = 0,$ which will be the tangent line slope:

$\frac{dy}{dx} \text{\textbar} _{x=y=0} = -1 + \frac{\exp(0+0)}{-0*\exp(0+0)+\cos(0+0)} = -1 + \frac{1}{-0*1 + 1} = -1 + 1 = 0.$

Therefore the equation of the tangent line is $y = 0.$