Let f(x)=9sinx/2sinx+4cosx
Then f'(x)=________ .
The equation of the tangent line to y=f(x) at a= pi/2 can be written in the form y=mx+b where
m= _____and
b=_____
doesn't look too bad as soon as you let us know whether the denominator is 2sinx or 2sinx+4cosx
(do you see how important brackets are?)
To find the derivative of the function f(x), we can use the quotient rule. The quotient rule states that for two functions u(x) and v(x), the derivative of u(x)/v(x) is given by:
f'(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))^2
In this case, u(x) = 9sin(x) and v(x) = 2sin(x) + 4cos(x). Let's calculate the derivatives of u(x) and v(x):
u'(x) = 9cos(x)
v'(x) = 2cos(x) - 4sin(x)
Now we can substitute these values into the quotient rule formula:
f'(x) = (9cos(x) * (2sin(x) + 4cos(x)) - 9sin(x) * (2cos(x) - 4sin(x))) / (2sin(x) + 4cos(x))^2
Simplifying this expression further will give us the derivative of f(x).
Now, to find the equation of the tangent line to y = f(x) at a = π/2, we need to find the slope (m) and the y-intercept (b) of the tangent line.
To find the slope, we substitute the value of a = π/2 into f'(x):
m = f'(π/2)
To find the y-intercept, we substitute the value of a = π/2 into f(x):
b = f(π/2)
Finally, we can write the equation of the tangent line in the form y = mx + b, using the values of m and b that we have found.