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# differentiate inverse tan watch

1. How would i differentiate y = tan(^-1) (1/X)

ive done tany = 1/x

ive got my answer down to (1+x^2)^-1

when it should be -(1+x^2)^-1

any help appreciated
x
2. do you mean ?
3. (Original post by zulu)
How would i differentiate y = tan(^-1) (1/X)

ive done tany = 1/x

ive got my answer down to (1+x^2)^-1

when it should be -(1+x^2)^-1

any help appreciated
x
Personally i would do the chain rule with t=1/x. the differential of tan^-1(t) you may already know and if not its in the formula book.
5. y=arctanx

x=tany
dx/dy = sec^2 x

Sec^2x=tan^2x + 1

dx/dy=tan^2x + 1

dy/dx=1/(tan^2x + 1)=1/(x^2 + 1)

I get the same answer as you.
6. (Original post by zulu)

ive got my answer down to (1+x^2)^-1
That's correct
7. Please note that in the question the op has set, it is tan^-1(1/x). The above two posts show it for tan^-1(x). The answer book has it right. The way to do this is by the chain rule.
8. In case anyone still needs the solution;

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

differentiate implicitly:
d/dx (cot y) = d/dx (x)
-cosec^2(y) dy/dx = 1
dy/dx = 1/(-cosec^2(x)) = 1/(-cot^2(y)-1)

since we know cot y = x
dy/dx becomes 1/(-x^2 - 1) which is the same as 1/-(x^2 + 1) or -(1+x^2)^-1

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