f(x)=4sinx/1+cosx
find f'(x)
this is what i got and i dunno what i did to get it wrong
(4cos(x)+ 4cosx^2-4sin(x)^2)/cosx62
looks like a ± error, and your denominator is wrong
my first line of the derivative was
(1+cosx)(4cosx) - 4sinx(-sinx))/(1+cosx)^2 using the quotient rule
expanding :
(4cosx + 4cos^2x + 4sin^2x)/(1+cosx)^2
= 4(cosx + 1)/(1+cosx)^2 , remember sin^2x+cos^2 = 1
= 4/(1+cosx)
To find the derivative of f(x) = 4sinx/(1+cosx), you can use the quotient rule. The quotient rule states that if you have a function of the form g(x)/h(x), where g(x) and h(x) are differentiable functions, the derivative of f(x) is given by:
f'(x) = (g'(x)h(x) - g(x)h'(x)) / h(x)^2
In this case, g(x) = 4sinx and h(x) = 1 + cosx.
Now, let's find the derivatives of g(x) and h(x) individually:
g'(x) = d/dx (4sinx) = 4cosx
h'(x) = d/dx (1 + cosx) = -sinx
Substituting these values into the quotient rule formula, we get:
f'(x) = (4cosx * (1 + cosx) - 4sinx * (-sinx)) / (1 + cosx)^2
Expanding the numerator, we obtain:
f'(x) = (4cosx + 4cos^2x + 4sin^2x) / (1 + cosx)^2
Now, note that cos^2x + sin^2x simplifies to 1 using the Pythagorean identity. Substituting this simplification, we have:
f'(x) = (4cosx + 4) / (1 + cosx)^2
Therefore, the derivative of f(x) is f'(x) = (4cosx + 4) / (1 + cosx)^2.
It seems that in the numerator, you expanded cosx^2 incorrectly as cosx^2 instead of cos^2x. Also, the denominator should not be cosx62; it should be (1 + cosx)^2.