The derivative of (tanx)^-1?

they are not same. (tan x)^-1 is reciprocal of tan x and tan^1- x is inverse function of tan x

(tan x)^-1 = 1/tanx = cot x

the derivative of cot x = -cosec^2(x)

this can be derived using cotx = cos x / sinx

now cot(x+h) = cos(x+h)/sin (x+h)

cot (x+h) - cot(x) = (-cos x sin (x+h) + sin x cos (x+h))/sin x sin(x+h)

= -sin h/(sin x sin (x+h))

(cot (x+h) - cot x)/h = - (sin h/h)/sin x sin (x+h)

as h->0 left hand side = df/dx and rhs = - 1/ sin^2 x = - cosec^2 x

but

tan^-1(x) = arctan(x)

derivative of arctan x = 1/(1+x^2)

this can be derived but you did not ask for it

Derivative Tan-1

Tan-1 Derivative

Rewrite as d/dx (1+tanx)^(-1) By the chain rule we have -1(1+tanx)^(-2) * d/dx(1+tanx) = -1(1+ tanx)^(-2) * sec^2(x) = -sec^2(x)/(1+tanx)^(2)

RE:

The derivative of (tanx)^-1?

Could someone help me find the derivative of (tanx)^-1? Does finding it have anything to do with tan^-1(x)? Thanks.

-1 times (tanx)^-2 times (sec x) times (tan x)

simplify this

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-1/(cos(x)+sin(x))^2 Scroll down to derivative and select choose; show steps.

No. Tan^-1(x) (also called arctan(x)) is a different thing.

(tan(x))^-1 = 1/(tan(x)) = 1/((sin(x))/(cos(x))) =

cos(x)/sin(x)

Now apply the quotient rule for derivatives and you're all done.

Doug

No no no! ALL WRONG!!

1 / (1 + x^2) is the definition of d/dx(arctan(x)).

Based on what the first guy said the answer you should get is -1/(sin^2(x)).