Final answer:
The equation of the tangent line to the function f(x) = 9e^xcos(x) at the point (0,9) is found by computing the function's derivative and evaluating it at x = 0 to get the slope of 9. Using the point-slope form, the equation of the tangent line simplifies to y = 9 + 9x.
Step-by-step explanation:
To find the equation of the tangent line to the graph of the function f(x) = 9excos(x) at the specified point (0,9), we first need to compute the derivative of the function to find the slope of the tangent. The derivative, f'(x), can be found using the product rule and the chain rule.
The product rule states that the derivative of two functions multiplied together, u(x)v(x), is u'(x)v(x) + u(x)v'(x). Applying this to f(x), we get:
f'(x) = d/dx(9excos(x)) = 9excos(x) - 9exsin(x)
At the point x = 0, the derivative simplifies to:
f'(0) = 9e0cos(0) - 9e0sin(0) = 9
Now, using the point-slope form of the line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point, the equation of the tangent line is:
y - 9 = 9(x - 0)
Which simplifies to the equation of the tangent line:
y = 9 + 9x