Question

find the derivative of the trigonometric function….
y=x+cotx

Answer 1

Hmm so you'll want to change cotangent to it's sine and cosine equivalent, and then apply the quotient rule :)

Remember the identity for cotangent?

Answer 2

how would i do that?...

Answer 3

the derivative of cotx is -csc^2(x) and you should know the derivative of x :)

Answer 4

did you get it ?

Answer 5

\[\cot(x)=\frac{ \cos(x) }{ \sin(x) }\]
\[\frac{ d }{ dx }\cot(x)=\frac{ -\sin^{2}(x)-\cos^{2}(x) }{ \sin^{2}(x) }=\frac{ -1 }{ \sin^{2}(x) }=-\csc^{2}(x)\]Now
\[\frac{ d }{ dx }y=1-\csc^{2}(x)=-\cot^{2}(x)\]

Answer 6

^ proper way to do it

Answer 7

^^ the way I do it because I don't remember d/dx of cot x :">