d/dx (sin^3 x) = d/dx (sinx sinx sinx)
use the chain rule
y = sin^3x
y' = 3 sin^2x cosx
d/dx (sinx sinx sinx)
=sin x d/dx(sin x sin x) + (sin x sin x) d/dx(sin x)
= sin x d/dx(sin x sin x) + (sin x sin x) cos x
= sin x[2sin x cos x] + sin^2 x cos x
= 2 sin^2 x cos x + sin^2 x cos x
= 3 sin^2 x cos x
which we knew anyway
To differentiate the function f(x) = sin^3(x), we can use the chain rule of differentiation.
The chain rule states that if we have a function g(x) being raised to a power n, we can differentiate it by taking the derivative of the inner function g(x) and multiplying it by n*g(x)^(n-1).
In this case, our inner function g(x) is sin(x), and it is being raised to the power of 3.
Let's differentiate step-by-step:
Step 1: Compute the derivative of the inner function g(x) = sin(x) using the chain rule:
d/dx (sin(x)) = cos(x)
Step 2: Apply the chain rule to the function f(x) = sin^3(x):
d/dx (sin^3(x)) = 3*sin^2(x) * d/dx(sin(x))
Step 3: Substitute the value we obtained from Step 1 into Step 2:
d/dx (sin^3(x)) = 3*sin^2(x) * cos(x)
So, d/dx (sin^3(x)) = 3*sin^2(x) * cos(x).
To find the derivative of the function f(x) = sin^3 x, we can apply the chain rule. The chain rule states that if we have a function g(u) and another function f(x), then the derivative of their composition is given by (g'(u) * f'(x)).
So, let's break down the expression f(x) = sin^3 x into the composition of two functions g(u) = u^3 and f(x) = sin x.
First, we differentiate g(u) = u^3 with respect to u. This is a simple power rule, which states that if we have a function of the form f(u) = u^n, then its derivative is given by f'(u) = n*u^(n-1). In this case, since n = 3, we have g'(u) = 3*u^(3-1) = 3*u^2.
Next, we differentiate f(x) = sin x with respect to x. The derivative of sin x is cos x, which can be obtained by applying the derivative of trigonometric functions.
Finally, we can use the chain rule to get the derivative of f(x) = sin^3 x. The derivative is given by f'(x) = g'(u) * f'(x). In this case, u = sin x and we already found that g'(u) = 3*u^2 and f'(x) = cos x. Substituting these values, we have f'(x) = 3*(sin x)^2 * cos x.
Thus, the derivative of sin^3 x with respect to x is 3*(sin x)^2 * cos x.