Question

Can someone explain how to find the derivative of this?

f(x) = cot^3(2x^3).

I know it equals -18x^2cot^2(2x^3)csc^2(2x^3), but I'm not sure why. I think it involves the chain rule, but I'm fuzzy on it.

Answer 1

\[f(x)=\cot(x) \rightarrow f \prime(x)=-\csc ^{2}(x)\]
\[f(x)=\cot ^{3}(x) \rightarrow f \prime(x)=-3*csc ^2(x)*cot ^{2}(x)\]
\[f(x)=\cot(2x ^{3}) \rightarrow f \prime(x)=-csc ^2(2x ^{3})*(6x ^{2})\]

or, more generally:
\[f(x)=\cot ^{n}(x) \rightarrow f \prime(x)=-n*csc ^2(x)*cot ^{n-1}(x)\]
\[f(x)=\cot(g(x)) \rightarrow f \prime(x)=-csc ^2(g(x))*(g \prime (x))\]
\[f(x)=\cot^{n}(g(x)) \rightarrow f \prime(x)=-n*csc ^2(g(x))*(g \prime (x))*cot ^{n-1}(g(x))\]

Answer 2

So... you take the derivative of the whole thing (3cot^2(2x^3)) and multiply it by the derivative of the inside (6x^2) and multiply that by -csc^2(2x^3). I'm not sure where the last part comes from. I realize that it's the derivative of cot(x), but the original function was cot^3, not cot^1. I'm having trouble seeing the steps. Reviewing it again...

Answer 3

well, you already (correctly) stated that \[f(x)=cot ^{3}(2x^{3}) \rightarrow f \prime (x)=-18x^{2}*cot^{2}(2x^{3})*csc^{2}(2x^{3})\]i was trying to show you the middle steps

Answer 4

not sure what "last part" you mean. mutliplying those three terms together should give you the correct function.

Answer 5

I meant... I didn't understand why we have to multiply \[3\cot^2(2x^3)(6x^2)\] by \[-\csc^2(2x^3)\] but I think I understand now. It's because we have to use the chain rule again and \[\cot^3(x)\] is the same as writing \[(\cot(x))^3\]