Question

Derivative of cos(a^3+x^3)

Answer 1

With respect to x, I suppose?

Answer 2

It's y= cos(a^3+x^3)

Answer 3

You have to apply the chain rule:\[\frac{d}{dx}\left [ \cos(a^3+x^3) \right ]=-3x^2\sin(a^3+x^3).\]Do you want to know how I got that?

Answer 4

Yes, please

Answer 5

Given two functions,\[f(x)\]and\[g(x),\]the derivative of\[f(g(x))\]can be found by applying the chain rule. That is,\[\frac{d}{dx}\left [ f(g(x)) \right ]=f'(g(x))g'(x).\]

In this example, you can see that\[f(x)=\cos x\]and\[g(x)=a^3+x^3,\]thus\[f(g(x))=cos(a^3+x^3),\]which is your question. To find its derivative, you have to know what\[f'(x)\]and\[g'(x)\]are. Namely,\[f'(x)=-\sin x\]\[g'(x)=3x^2.\]From the chain rule above, we can conclude that the derivative of\[cos(a^3+x^3)\]is\[-3x^2\sin(a^3+x^3).\]Do you have any questions?

Answer 6

How did you get the 3x^2] by itself. Where did the 'a" go?

Answer 7

I'm assuming that the 'a' here is a constant. Remember that the derivative of a constant will always be zero. That is,\[\frac{d}{dx}\left [ c \right ]=0.\]

Answer 8

I see. Thank You for explaining

Answer 9

That's why I asked you if you were differentiating with respect to 'x', because if 'a' is a variable here, then the function's derivative would end up being a partial differential.

No, you're welcome!