f(x)=sin3x, find f^(79) and f^(120)?
What am I suppose to do?
Is f^ supposed to be the first derivative if f?
Is the f(x) function sin^3(x) or sin(3x) ?
it is f(x)= sin(3x) and yes it is suppose to be the derivative
y = sin t
dy/dx = dt/dx cos t
so if
y = sin 3x
dy/dx = d/dx(3 x) cos 3 x
= 3 cos 3x
so if x = 79 for example
dy/dx = 3 cos 3*79 = 3 cos 237
by the way I am surprised that it is in fact not
y = sin^3 x
in which case you would get
dy/dx = 3 sin^2x cos x
To find the derivative of a function, you can use the power rule, chain rule, and trigonometric identities.
In this case, you are asked to find the 79th and 120th derivatives of the function f(x) = sin(3x).
To find the derivative of f(x), you can apply the chain rule. The chain rule states that if you have a function g(u) composed with another function f(x), the derivative of g(f(x)) can be found by multiplying the derivative of g(u) with the derivative of f(x) with respect to x.
In this case, let's start by finding the derivative of sin(3x). The derivative of sin(x) with respect to x is cos(x), so applying the chain rule, the derivative of sin(3x) with respect to x is cos(3x) times the derivative of 3x with respect to x, which is 3.
Therefore, the first derivative of f(x) = sin(3x) is f'(x) = 3cos(3x).
To find higher-order derivatives (in this case, the 79th and 120th derivatives), you can use the same process repeatedly.
To find the 79th derivative, take the derivative of f'(x) = 3cos(3x) 78 more times. Each time you apply the derivative, multiply by the derivative of 3cos(3x) with respect to x, which is -9sin(3x). After applying the derivative 78 times, you will be left with (-9)^39 * sin(3x).
Therefore, the 79th derivative of f(x) = sin(3x) is f^(79)(x) = (-9)^39 * sin(3x).
Similarly, to find the 120th derivative, take the derivative of f'(x) = 3cos(3x) 119 more times. After applying the derivative 119 times, you will be left with (-9)^59 * cos(3x).
Therefore, the 120th derivative of f(x) = sin(3x) is f^(120)(x) = (-9)^59 * cos(3x).