<s>[INST] Using the chain rule for cos and sin functions I am having issues with derivatives containing chain rules. I know there is multiple threads already but after reading a few, I still find myself confused. I also checked the actual answer following a step by step website without success.
Derivate: $$h(x)=\sin ( x^6 - cos^3 x^2)$$
Now I have $sin = f(x)$ and $( x^6 - cos^3 x^2) = g(x)$ so i have a $f(g(x))$ form
The general formula says $f(g(x))$  = $f'(g(x))*g'(x)$
First, let's derivate the first part using --> $sin f(x) = cos f(x) * f'(x)$
$$cos (x^6-cos^3x^2)*(6x^5 - ??)$$
I know that $cos^3 x$ is $(cos x)^3$ so that means $((cos x^2)^3)'= 3(cos x^2)^2*-sinx^2*2x$
So we have
$$cos (x^6-cos^3x^2)*(6x^5-(6x(cos x^2)^2*-sinx^2)))$$
Simplified
$$cos (x^6-cos^3x^2)(6x^5-6xcos^2 x^2*sinx^2)$$
Where i'm confused : The rule says that we need to do f'(g(x))*g'(x), this would mean that I would need to do:
$$(sin(x^6-cos^3x^2))'*(x^6-cos^3x^2)'$$
Which would make an even bigger answer
$$cos (x^6-cos^3x^2)*(6x^5-6xcos^2 x^2*sinx^2) * (6x^5-6xcos^2 x^2*sinx^2)$$
And then I don't understand.
Also, it seems that i have done a mistake because the actual answer is the following (I have a $-$ where it's suppose to be a $+$)
Real answer : $$[cos (x^6-cos^3x^2)](6x^5+6xcos^2 x^2sinx^2)$$
My answer : $$cos (x^6-cos^3x^2)(6x^5-6xcos^2 x^2*sinx^2)$$
 [/INST] $f'(g(x))\cdot g'(x)$ means, in your case, $\sin'(x^6-\cos^3{x^2})\cdot (x^6-\cos^3(x^2))'$ and not what you have written. This would give the result to be $\cos(x^6-\cos^3(x^2))\cdot (6x^5-(3\cos^2(x^2)(-\sin (x^2))(2x))$ which equals the real answer given, i.e. $\cos(x^6-\cos^3(x^2))\cdot(6x^5+6x\cos^2(x^2)\sin(x^2))$
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