If f(x)=4sinx/1+cosx
then f'(x)=_____?
I think that the derivative of 4sinx would be 4cos and cosx would be sinx but I don't know where to go from there.
put u=4sin(x), and v=(1+cos(x)),
then use the quotient rule:
d(u/v)= (v du/dx - u dv/dx) / v2
Sorry, I didn't see the comment the first time,
d(cos(x))/dx = -sin(x)
To find the derivative of f(x) = 4sinx / (1 + cosx), we need to use the quotient rule.
The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), then the derivative of f(x) is given by:
f'(x) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
Applying the quotient rule to f(x) = 4sinx / (1 + cosx), we get:
f'(x) = [(1 + cosx) * (4cosx) - (4sinx) * (-sinx)] / (1 + cosx)^2
Simplifying further,
f'(x) = (4cos^2x + 4cosx + 4sin^2x) / (1 + cosx)^2
Now, using the trigonometric identity sin^2x + cos^2x = 1, we can rewrite the numerator as:
f'(x) = 4(1 + cosx) / (1 + cosx)^2
Finally, canceling out the common factor of (1 + cosx) from the numerator and the denominator, we get:
f'(x) = 4 / (1 + cosx)
To find the derivative of the given function f(x) = 4sinx/(1+cosx), we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), where g(x) and h(x) are differentiable functions, then the derivative of f(x) is given by:
f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2
Let's apply the quotient rule to find the derivative of f(x) step by step.
Step 1: Find the derivative of g(x)
g(x) = 4sinx
To find the derivative of sin(x), we know that the derivative of sin(x) is cos(x). Therefore, g'(x) = 4cos(x).
Step 2: Find the derivative of h(x)
h(x) = 1 + cos(x)
To find the derivative of cos(x), we know that the derivative of cos(x) is -sin(x). Therefore, h'(x) = -sin(x).
Step 3: Apply the quotient rule
Using the quotient rule formula, we can substitute g'(x) = 4cos(x), g(x) = 4sin(x), h'(x) = -sin(x), and (h(x))^2 = (1 + cos(x))^2 into the formula for f'(x):
f'(x) = (4cos(x)(1 + cos(x)) - 4sin(x)(-sin(x))) / (1 + cos(x))^2
Simplifying further:
f'(x) = (4cos(x) + 4cos^2(x) + 4sin^2(x)) / (1 + cos(x))^2
Using the identity sin^2(x) + cos^2(x) = 1, we can simplify the expression:
f'(x) = (4cos(x) + 4) / (1 + cos(x))^2
Therefore, the derivative of f(x) = 4sinx/(1+cosx) is f'(x) = (4cos(x) + 4) / (1 + cos(x))^2.