Find a equation of the tangent line at the point specified.

Question

y=e^xcosx

x=0

Answer 1

The equation of a line can be specified as y = mx + b where m is the slope and b is the y-intercept. The y-intercept is when x = 0 which just happens to be the x value of the point they are asking about. What is the y value of that point? To find out, plug in 0 into the function. So f(0) = e^(0)cos(0) = e^0*1 = e⁰ = 1. That is the "b" in y = mx + b. The m (slope of the tangent line) is the derivative of f(x) at x = 0. To determine f ' we must use the chain rule.

f ' (x) = e^xcos(x)(xcos(x)) '

to find (xcos(x)) ' we must use the product rule. (xcos(x)) ' = x'cos(x) + xcos(x) ' = 1*cos(x) + x(-sin(x)) = cos(x) - xsin(x)

So f ' (x) = e^xcos(x)(cos(x) - xsin(x))

And f ' (0) = e^0cos(0)(cos(0) - 0sin(0)) = e⁰(1 - 0) = 1*1 = 1

So since m = 1 and b = 1, the equation is y = x + 1