How do I find slope of the tangent to the curve y=(e^x+e^-x)/2 at point (0,1)?
To find the slope of the tangent to the curve at a given point, you need to take the derivative of the function representing the curve at that point. In this case, the function representing the curve is y = (e^x + e^-x) / 2.
To find the derivative, you can use the rules of differentiation. The derivative of e^x with respect to x is e^x, and the derivative of e^-x with respect to x is -e^-x. The constant factor 1/2 can be ignored while taking the derivative.
So, the derivative of y with respect to x, denoted as dy/dx or y', is:
dy/dx = (d/dx)(e^x + e^-x) / 2
= d/dx (e^x) + d/dx (e^-x) / 2
= e^x - e^-x / 2
Now that we have the derivative dy/dx, we can substitute the x-coordinate of the given point (0, 1) into the derivative expression to find the slope of the tangent at that point. So, when x = 0:
slope = dy/dx |(x=0)
= (e^0 - e^0) / 2
= (1 - 1) / 2
= 0 / 2
= 0
Therefore, the slope of the tangent to the curve y = (e^x + e^-x) / 2 at the point (0, 1) is 0.