How do you differentiate $y = a r c cot (\frac{x}{5})$ $y=arc cot(x/5)$ ?

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How do you differentiate $y = a r c cot (\frac{x}{5})$ $y=arc cot(x/5)$ ?

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Answer 1 · 2017-07-12T00:53:03

I got

$\frac{d y}{d x} = - \frac{1}{5} ⎛ ⎜ ⎜ ⎝ \frac{1}{1 + {(\frac{x}{5})}^{2}} ⎞ ⎟ ⎟ ⎠$ $(dy)/(dx) = -1/5 (1/(1 + (x/5)^2))$

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

Instead, I'll rewrite this as

$cot y = \frac{x}{5}$ $coty = x/5$

Then, by implicit differentiation:

$- {csc}^{2} y \frac{d y}{d x} = \frac{1}{5}$ $-csc^2y (dy)/(dx) = 1/5$

Therefore:

$\frac{d y}{d x} = - \frac{1}{5 {csc}^{2} y}$ $(dy)/(dx) = -1/(5csc^2y)$

And since we defined $y = a r c cot (\frac{x}{5})$ $y = arc cot(x/5)$ , we can use the identity

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

to get

$\frac{d y}{d x} = - \frac{1}{5} \frac{1}{1 + {cot}^{2} y}$ $color(blue)((dy)/(dx)) = -1/5 1/(1 + cot^2y)$

$= - \frac{1}{5} ⎛ ⎜ ⎜ ⎝ \frac{1}{1 + {(\frac{x}{5})}^{2}} ⎞ ⎟ ⎟ ⎠$ $= color(blue)(-1/5 (1/(1 + (x/5)^2)))$

If you did it using the actual derivative of $a r c cos u$ $arc cosu$ , you would still get

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

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

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How do you differentiate $y = a r c cot (\frac{x}{5})$ $y=arc cot(x/5)$ ?

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