Question

If m(x)=sin²(x), then m'(x)=?​

Answer 1

To find the derivative of m(x) = sin²(x), we can use the chain rule:

m'(x) = 2sin(x)cos(x)

To derive this, we can start by writing m(x) as a composition of two functions: m(x) = f(g(x)), where f(u) = u² and g(x) = sin(x). Then, we can apply the chain rule:

m'(x) = f'(g(x))g'(x)
= 2g(x)cos(x)
= 2sin(x)cos(x)

where we have used the fact that f'(u) = 2u for any function f(u) = u². Thus, the derivative of m(x) = sin²(x) is m'(x) = 2sin(x)cos(x).

Answer 2

To find the derivative of m(x) = sin²(x), we can use the chain rule. The chain rule tells us that if we have a function g(x) inside of another function f(x), then the derivative of f(g(x)) is f'(g(x)) times g'(x).

So in this case, we have f(x) = sin²(x) and g(x) = sin(x). We know the derivative of sin(x) is cos(x), so g'(x) = cos(x).

To find f'(g(x)), we need to take the derivative of sin²(x) with respect to sin(x). This is a little bit tricky, but we can use the chain rule again! Let u = sin(x), so that f(x) = u². Then f'(x) = 2u * du/dx, where du/dx is the derivative of sin(x), which we already know is cos(x).

Putting this all together, we get:

m'(x) = f'(g(x)) * g'(x) = 2sin(x) * cos(x) = sin(2x)

Therefore, m'(x) = sin(2x).