What is the equation of the line tangent to #f(x)=y=e^x sin^2x# at #x=sqrtpi#?

Question

Paisley Johnson · Accepted Answer

The equation is approximately:
#y = 3.34x - 0.27# To start, we need to determine #f'(x)#, so that we know what the slope of #f(x)# is at any point, #x#. #f'(x) = d/dx f(x) = d/dx e^x sin^2(x)# using the product rule: #f'(x) = (d/dx e^x)sin^2(x) + e^x(d/dx sin^2(x))# These are standard derivatives: #d/dx e^x = e^x# #d/dx sin^2(x) = 2sin(x)cos(x)# So our derivative becomes: #f'(x) =e^x sin(x) (sin(x) + 2cos(x))# Inserting the given #x# value, the slope at #sqrt(pi)# is: #f'(sqrt(pi)) = e^(sqrt(pi))sin(sqrt(pi)) (sin(sqrt(pi)) + 2cos(sqrt(pi)))# This is the slope of our line at the point #x= sqrt(pi)#. We can then determine the y intercept by setting: #y = mx + b# #m = f'(sqrt(pi))# #y = f(sqrt(pi))# This gives us the non-simplified equation for our line: #f(sqrt(pi)) = (e^(sqrt(pi))sin(sqrt(pi)) (sin(sqrt(pi)) + 2cos(sqrt(pi))))x + b# #e^(sqrt(pi)) sin^2(sqrt(pi)) = (e^(sqrt(pi))sin(sqrt(pi)) (sin(sqrt(pi)) + 2cos(sqrt(pi))))x + b# Solving for b, we end up with the annoyingly complicated formula: #b = e^(sqrt(pi)) sin sqrt(pi)[ sin sqrt(pi) - sqrt(pi)(sin(sqrt(pi)) + 2 cos (sqrt(pi))]# So our line ends up being: #y = [e^(sqrt(pi))sin(sqrt(pi)) (sin(sqrt(pi)) + 2cos(sqrt(pi)))]x + e^(sqrt(pi)) sin sqrt(pi)[ sin sqrt(pi) - sqrt(pi)(sin(sqrt(pi)) + 2 cos (sqrt(pi))]# If we actually calculate what these annoyingly large coefficients equate to, we end up with the approximate line: #y = 3.34x - 0.27#

Charles Anctil · Answer

undefined To find the equation of the line tangent to the function f(x) = y = e^x sin^2x at x = sqrt(pi), we need to find the slope of the tangent line and the point of tangency.

First, let's find the derivative of f(x) with respect to x:
f'(x) = (d/dx) (e^x sin^2x)
Using the product rule and chain rule, we get:
f'(x) = e^x * 2sinx * cosx + e^x * sin^2x

Now, let's evaluate f'(x) at x = sqrt(pi):
f'(sqrt(pi)) = e^(sqrt(pi)) * 2sin(sqrt(pi)) * cos(sqrt(pi)) + e^(sqrt(pi)) * sin^2(sqrt(pi))

Next, let's find the y-coordinate at x = sqrt(pi) by substituting it into the original function:
f(sqrt(pi)) = e^(sqrt(pi)) * sin^2(sqrt(pi))

Now, we have the slope of the tangent line (f'(sqrt(pi))) and the point of tangency (sqrt(pi), f(sqrt(pi))). Using the point-slope form of a line, the equation of the tangent line is:
y - f(sqrt(pi)) = f'(sqrt(pi)) * (x - sqrt(pi))

Simplifying the equation will give you the final answer.