Can someone explain to me why the derivative of sinx^2 is y'=2xcos (x^2)???
d/dx sin(a) = da/dx cos(a)
the derivative of the sine is the cosine
but the argument may also have a derivative
Review the chain rule
d/dx f(u(x)) = df/du * du/dx
Certainly! To understand why the derivative of sin(x^2) is y' = 2x * cos(x^2), we need to use the chain rule.
The chain rule is a rule in calculus that allows us to find the derivative of a composite function. In this case, the function f(x) = sin(x^2) can be thought of as the composition of two functions: the outer function g(x) = sin(x) and the inner function h(x) = x^2.
To apply the chain rule, we need to find the derivative of the outer function (g(x)) and the derivative of the inner function (h(x)), and then multiply them. Let's break it down step by step:
1. The derivative of the outer function g(x) = sin(x) is calculated as g'(x) = cos(x). This is a well-known derivative of sine function.
2. The derivative of the inner function h(x) = x^2 is calculated as h'(x) = 2x. This is the power rule for the derivative of x^n.
Now, using the chain rule formula, the derivative of f(x) = g(h(x)) is given by:
f'(x) = g'(h(x)) * h'(x)
Substituting the derivatives we found earlier:
f'(x) = cos(x^2) * 2x
As you can see, the derivative of sin(x^2) is indeed y' = 2x * cos(x^2).
So, to get this result, we used the chain rule, which involves finding the derivative of the outer function (sin(x)) and the derivative of the inner function (x^2), and then multiplying them together.