what is the derivative of y=4secx^2=3cotx ?
Is the second = supposed to be a minus? In other words, is the function
y = 4secx^2 -3cotx ?
Also, do you mean (secx)^2, or sec(x^2) ?
Sorry about that. y=4sec(x^2)-3cotx
The derivative of secx is tanx*secx.
The derivative of cotx is -csc^2x
Treat sec*(x^2) as a function f of a function g(x). f(g) = sec g; g(x) = x^2
d/dx[sec(g(x))]= d/dg(secg)*dg/x
d/dx[sec(x^2)] = tanx^2)*sec(x^2)*2x
Now add them all up, and include the constant coefficients
dy/dx = 8x tan(x^2)*sec(x^2) + 3csc^2 x
To find the derivative of the given function y = 4sec^2(x) - 3cot(x), you can follow these steps:
Step 1: Rewrite the function using trigonometric identities.
The function y can be rewritten as y = 4 * (1/cos^2(x)) - 3 * (cos(x)/sin(x)).
Step 2: Simplify the function.
Simplifying further, y = 4/cos^2(x) - 3cos(x)/sin(x).
Step 3: Apply the quotient rule to find the derivative.
The quotient rule states that if you have a function in the form u(x)/v(x), its derivative can be calculated as (v(x) * u'(x) - u(x) * v'(x)) / (v(x))^2.
Let's apply this rule:
u(x) = 4
v(x) = cos^2(x) - 3sin(x)cos(x)
Now, let's find the derivatives of u(x) and v(x):
u'(x) = 0 (the derivative of a constant is zero)
v'(x) = -2sin(x)cos(x) - 3cos^2(x) + 3sin^2(x)
Now, substitute the values into the quotient rule:
y' = [v(x) * u'(x) - u(x)*v'(x)] / (v(x))^2
= [cos^2(x) - 3sin(x)cos(x) * 0 - 4 * (-2sin(x)cos(x) - 3cos^2(x) + 3sin^2(x))] / (cos^2(x) - 3sin(x)cos(x))^2
Step 4: Simplify the expression.
Simplifying further, y' = [12sin^2(x)cos(x) - 6cos^3(x)] / (cos^2(x) - 3sin(x)cos(x))^2
Therefore, the derivative of y = 4sec^2(x) - 3cot(x) with respect to x is y' = [12sin^2(x)cos(x) - 6cos^3(x)] / (cos^2(x) - 3sin(x)cos(x))^2.