find d/dx (sin^(21)x *cos21x)
If that's sin(21x)*cos(21x) then
f = 1/2 sin(42x)
f' = 21cos(42x)
or, directly,
f' = 21cos(21x)cos(21x) - 21sin(21x)sin(21x)
= 21cos221x - 21sin221x
On the other hand, if that's
f = sin21x*cos21x, then we have
f' = 21sin20x cos22x - 21sin22x cos20x
= 21sin20x cos20x (cos2x - sin2x)
= 21sin20x cos20x cos2x
To find the derivative of the function f(x) = sin^(21)(x) * cos(21x), we can use the product rule, which states that the derivative of a product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function.
Let's break down the function into two parts:
f(x) = u(x) * v(x)
where u(x) = sin^(21)(x) and v(x) = cos(21x).
Now, let's find the derivatives of u(x) and v(x):
Derivative of u(x):
To find the derivative of u(x), we need to apply the chain rule since we have the sine function raised to the power of 21.
Let's define U(x) = sin(x). Therefore, u(x) = U(x)^21.
Using the chain rule, the derivative of u(x) is given by:
du/dx = 21 * U(x)^(21-1) * dU/dx
The derivative of sin(x) is cos(x), so dU/dx = cos(x).
Substituting this back into the equation, we get:
du/dx = 21 * sin(x)^(21-1) * cos(x)
du/dx = 21 * sin^(20)(x) * cos(x)
Derivative of v(x):
To find the derivative of v(x), we differentiate cos(21x) using the chain rule.
dv/dx = -21 * sin(21x)
Now, we can calculate the derivative of f(x) using the product rule:
d/dx (u(x) * v(x)) = u(x) * dv/dx + v(x) * du/dx
d/dx (sin^(21)(x) * cos(21x)) = (sin^(20)(x) * cos(x)) * (-21 * sin(21x)) + (cos(21x)) * (21 * sin^(20)(x) * cos(x))
Simplifying the expression, we have:
d/dx (sin^(21)(x) * cos(21x)) = -21 * sin^(20)(x) * cos^2(x) * sin(21x) + 21 * sin^(20)(x) * cos^2(21x)
So, the derivative of the function f(x) = sin^(21)(x) * cos(21x) is:
d/dx (sin^(21)(x) * cos(21x)) = -21 * sin^(20)(x) * cos^2(x) * sin(21x) + 21 * sin^(20)(x) * cos^2(21x)