How do you differentiate $y=arc cot(x/5)$ ?

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Answer 1 · 2017-07-12T00:53:03

I got

$(dy)/(dx) = -1/5 (1/(1 + (x/5)^2))$

I'll assume you don't know that $d/(dx)[arc cot(u(x))] = -1/(1 + u^2)(du)/(dx)$ .

Instead, I'll rewrite this as

$coty = x/5$

$-csc^2y (dy)/(dx) = 1/5$

Therefore:

$(dy)/(dx) = -1/(5csc^2y)$

And since we defined $y = arc cot(x/5)$ , we can use the identity

$csc^2y = 1 + cot^2y$

to get

$color(blue)((dy)/(dx)) = -1/5 1/(1 + cot^2y)$

$= color(blue)(-1/5 (1/(1 + (x/5)^2)))$

If you did it using the actual derivative of $arc cosu$ , you would still get

$(dy)/(dx) = -1/(1 + underbrace((x/5)^2)_(u^2)) cdot d/(dx)[x/5]$

$= -1/5 (1/(1 + (x/5)^2))$

How do you differentiate y=arc cot(x/5)y=arc cot(x/5)?