differentiate
f(x)=(sinˆ2x)/(cosx)
if f = u/v,
f' = (u'v - uv')/v^2
so, whatcha got?
i am having problems with the trig identities that and how to solve with sin squared. i have the formula. that hasn't helped me with the identities.
To differentiate the function f(x) = (sin^2x)/(cosx), we can use the quotient rule. The quotient rule states that if you have a function of the form f(x) = g(x)/h(x), then the derivative of f(x) is given by the formula:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Let's apply this rule to differentiate f(x) = (sin^2x)/(cosx).
Step 1: Identify the functions g(x) and h(x).
In this case, g(x) = sin^2x and h(x) = cosx.
Step 2: Differentiate g(x) and h(x) separately.
To differentiate sin^2x, we can use the chain rule.
g'(x) = 2sinx * cosx
To differentiate cosx, we can use the sine rule.
h'(x) = -sinx
Step 3: Plug the derived values back into the quotient rule formula.
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= ((2sinx * cosx) * cosx - sin^2x * (-sinx)) / (cosx)^2
= (2sinx * cos^2x + sin^3x) / cos^2x
So, the derivative of f(x) = (sin^2x)/(cosx) is f'(x) = (2sinx * cos^2x + sin^3x) / cos^2x.