Prove that (cot x)' = - csc^2 x
- print Print
- list Cite
Expert Answers
calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
Let f(x) = cotx
Then we know that: cot x = cosx/sinx
==> f(x) = cosx/sinx
Let us differetiate using the chain rule:
f(x) = u/v such that:
u= cosx =-=> u' = -sin
v= sinx ==>...
(The entire section contains 104 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- Prove the identity: csc x - cot x = sinx/1+cosx
- 1 Educator Answer
- Prove: (1 + csc x)/sec x - cot x = cos x
- 1 Educator Answer
- Prove csc^2 x + sec^2 x = csc^2 x sec^2 xRS shows the 2 sets being multiplied.
- 1 Educator Answer
- Cot^2(x)/csc(x)-1=csc(x)+sen^2(x)+cos^2(x)
- 1 Educator Answer
- Prove that `(1 + tan x)/(1+cot x) = tan x`
- 2 Educator Answers
To prove that (cotx)' = -cosec^2 x
Proof:
LHS = (cotx)' = (cosx/sinx)'
(cotx)' = {(cosx)'(sinx)- (cosx)(sinx)'}/(sinx)^2, as (u(x)/v(x))' = {u'(x)v(x)-u(x)v'(x)}/(v(x))^2.
(cotx)' = {(-sinx)(sinx) -cosx)(cosx)}/(sinx).
(cotx)' = -(sinx^2x+cos^2x)/(sin^2x) .
(cotx)' = -1/sin^2x, as sin^2x+cos^2x = 1 is an identity.
(cotx)' = - cosec^2x.
So (cotx)' = - cosec^2x is proved.
We know that cosec x = 1/sin x and cot x = cos x/ sin x
If we have to calculate the derivative of cot x, we'll consider the quotient rule:
(cos x/sin x)' = (cos x)'*(sin x) - (cos x)*(sin x)'/(sin x)^2
We'll calculate the derivatives of the functions sine and cosine:
(cos x)' = -sin x
(sin x)' = cos x
(cot x)' = [(-sin x)*(sin x) - (cos x)^2]/(sin x)^2
(cot x)' = - [(sin x)^2 + (cos x)^2]/(sin x)^2
But, from the fundamental formula of trigonometry, we'll have:
[(sin x)^2 + (cos x)^2] = 1
(cot x)' = -1/(sin x)^2
But 1/sin x = csc x.
If we'll raise to square both sides, we'll get:
1/(sin x)^2 = -(csc x)^2
Student Answers