i don't seem to understand the chain rule: deriv. of sin x^2 is (cos x^2)(2x), but...

[DuctTapePro]

chain rule says the derivative of sin x^2 is

(cos x^2)(2x)

but the derivative of sin x is

cos x

so... sin x^2 = cos x^2

and

(cos x^2)(2x) = cos x^2 ???!!!!???!!!

 

[Deleted member 4993]

chain rule says the derivative of sin x^2 is

(cos x^2)(2x)

but the derivative of sin x is

cos x

so... sin x^2 = cos x^2

and

(cos x^2)(2x) = cos x^2 ???!!!!???!!!

Do you understand what is meant by F[g(x)]? It is a function of a function.

cos(x^2) is a function of x^2.

here:

\(\displaystyle \displaystyle{\dfrac{d}{dx}F[g(x)] = \dfrac{dF[g(x)]}{dg} * \dfrac{dg}{dx} }\)

 

[Steven G]

chain rule says the derivative of sin x^2 is

(cos x^2)(2x)

but the derivative of sin x is

cos x

so... sin x^2 = cos x^2

and

(cos x^2)(2x) = cos x^2 ???!!!!???!!!

The derivative of sin of any angle is cos of that same angle TIMES the derivative of the angle. This is the rule that will help you get these problems correct.
For the record, The derivative of cos of any angle is sin of that same angle TIMES the derivative of the angle.
The derivative of tan of any angle is sec² of that same angle TIMES the derivative of the angle.

The angle never changes!

 

[JeffM]

\(\displaystyle y = sin(x^2).\)

\(\displaystyle \text {Let } u = x^2 \implies \dfrac{du}{dx} = 2x\) No mystery there.

\(\displaystyle \therefore y = sin(u) \implies \dfrac{dy}{du} = cos(u).\) No mystery there.

And the chain rule says

\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} \implies\)

\(\displaystyle \dfrac{dy}{dx} = cos(u) * 2x = cos(x^2) * 2x \ \because \ u = x^2.\)