How do you find an equation of the tangent line to the curve at the given point #f(x) = sin(cosx) # and #x=pi/2#?

Question

Ezra Adams · Accepted Answer

Take the derivative.
Plug in your x-value.
Use the solution to write a new equation.

#f(pi/2)=sin(cos(pi/2))=sin(0) = 0# #f'(x) = -sinx(cos(cosx))# #f'(pi/2) = -sin(pi/2)(cos(cos(pi/2))# #=-(1)(cos(0))=-1=m# #y-y_1 = m(x-x_1)# #y-0=-1(x-pi/2)# #y=-x+pi/2#

Emily Ammerman · Answer

undefined To find the equation of the tangent line to the curve at the given point, we need to find the derivative of the function f(x) = sin(cosx) and evaluate it at x = pi/2. The derivative of f(x) can be found using the chain rule. The derivative of sin(cosx) with respect to x is -cos(cosx) * sinx. Evaluating this derivative at x = pi/2, we get -cos(cos(pi/2)) * sin(pi/2). Simplifying further, we have -cos(0) * 1, which equals -1. Therefore, the slope of the tangent line at x = pi/2 is -1.

To find the equation of the tangent line, we use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope. Plugging in the values x1 = pi/2, y1 = f(pi/2) = sin(cos(pi/2)) = sin(0) = 0, and m = -1, we have y - 0 = -1(x - pi/2). Simplifying this equation, we get y = -x + pi/2. Therefore, the equation of the tangent line to the curve f(x) = sin(cosx) at x = pi/2 is y = -x + pi/2.