How do you find the equation of the tangent line to the curve #y=2xsinx# at #(pi/2,pi)#?

Question

Charles Arocha · Accepted Answer

The equation of the tangent is #y=2x#

I thought it would be easier to make a little video rather than try and write out my explanation.

I hope it helps :)

Cameron Alexander · Answer

undefined To find the equation of the tangent line to the curve y=2xsinx at the point (pi/2, pi), we need to find the slope of the tangent line and then use the point-slope form of a linear equation.

To find the slope of the tangent line, we can use the derivative of the function y=2xsinx. Taking the derivative of y with respect to x, we get:

dy/dx = 2sinx + 2xcosx

Now, substitute the x-coordinate of the given point (pi/2) into the derivative to find the slope at that point:

dy/dx = 2sin(pi/2) + 2(pi/2)cos(pi/2)
       = 2(1) + 2(pi/2)(0)
       = 2

So, the slope of the tangent line at (pi/2, pi) is 2.

Using the point-slope form of a linear equation, we can write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values of the point (pi/2, pi) and the slope (2) into the equation, we get:

y - pi = 2(x - pi/2)

Simplifying the equation, we have:

y = 2x - pi + 2(pi/2)
y = 2x - pi + pi
y = 2x

Therefore, the equation of the tangent line to the curve y=2xsinx at the point (pi/2, pi) is y = 2x.