The derivative of
cot2(x) is
−2cot(x)csc2(x).
Solution. Let
F(x)=cot2(x),
f(u)=u2, and
g(x)=cot(x) such that
F(x)=f(g(x)). We will use the chain rule to determine the derivative of
cot2(x):
F′(x)=f′(g(x))g′(x).
Earlier, we saw
here that
g′(x)=dxdcot(x)=−csc2(x), and
f′(u)=2u. Therefore, we get:
f′(g(x))=2g(x)=2cot(x)andg′(x)=−csc2(x).
So, we have:
F′(x)=f′(g(x))g′(x)=2cot(x)⋅(−csc2(x))=−2cot(x)csc2(x).
Therefore, the derivative of
cot2(x) is
−2cot(x)csc2(x).