Let f(x)=9sinx/2sinx+4cosx.
Then 'f(x)= ((2sinx+4cosx)(9cosx)-(9sinx)(2cosx-4sinx))/(2sinx+4cosx)^2.
The equation of the tangent line to y=f(x) at a=2 can be written in the form y=mx+b where
m= 9and
b= ______.?????
To find the equation of the tangent line to y = f(x) at a = 2, we need to calculate the value of f(2) and the derivative of f(x) at x = 2.
First, let's find f(2). Substitute x = 2 into the function f(x):
f(2) = (9sin(2)) / (2sin(2) + 4cos(2))
Now, let's find the derivative of f(x):
f'(x) = [(2sin(x) + 4cos(x))(9cos(x)) - (9sin(x))(2cos(x) - 4sin(x))] / (2sin(x) + 4cos(x))^2
Next, substitute x = 2 into the derivative to find f'(2):
f'(2) = [(2sin(2) + 4cos(2))(9cos(2)) - (9sin(2))(2cos(2) - 4sin(2))] / (2sin(2) + 4cos(2))^2
Now, let's calculate f(2) and f'(2):
f(2) = (9sin(2)) / (2sin(2) + 4cos(2))
f'(2) = [(2sin(2) + 4cos(2))(9cos(2)) - (9sin(2))(2cos(2) - 4sin(2))] / (2sin(2) + 4cos(2))^2
After calculating these values, substitute m = 9 into the equation y = mx + b and solve for b:
f(2) = 9(2) + b
f(2) - 18 = b
Therefore, b = f(2) - 18. Substitute the value of f(2) into the equation to find the value of b.