Differentiate with respect to x

**HNCMATHS** · January 29th 2011, 04:54 PM

Hi everyone, which way should I approach this one....

Differentiate with respect to x

$Y = 3 arccot 2$

**skeeter** · January 29th 2011, 05:19 PM

Originally Posted by HNCMATHS

Hi everyone, which way should I approach this one....

Differentiate with respect to x

$Y = 3 arccot 2$

since $3arccot(2)$ is a constant, the derivative is 0.

if you meant $y = 3arccot(2x)$ , then ...

$\dfrac{d}{dx} [k \cdot arccot(u)] = -\dfrac{k}{1 + u^2} \cdot \dfrac{du}{dx}$

**HNCMATHS** · January 29th 2011, 05:21 PM

Originally Posted by skeeter

since $3\arctan(2)$ is a constant, the derivative is 0.

if you meant $y = 3\arctan(2x)$ , then ...

$\dfrac{d}{dx} [k\arctan{u}] = \dfrac{k}{1 + u^2} \cdot \dfrac{du}{dx}$

Hi thanks for the reply, however the problem I am working with is definately 3arccot 2t

**dwsmith** · January 29th 2011, 05:24 PM

Originally Posted by HNCMATHS

Hi everyone, which way should I approach this one....

Differentiate with respect to x

$Y = 3 arccot 2$

Well, you can either just accept that

$\displaystyle\frac{d}{dx}\left[\cot^{-1}(x)\right]=\frac{-1}{1+x^2}$

or

If $y=\cot^{-1}(x)$ , then $x=\cot(y)$

Now derive $x=\cot(y)$ .

$\displaystyle 1=-\csc^2(y)\frac{dy}{dx}$

$\displaystyle\frac{dy}{dx}=-\frac{1}{\csc^2(y)}$

$\csc=\sqrt{1+\cot^2}=\sqrt{1+x^2}$

$\displaystyle\frac{dy}{dx}=-\frac{1}{1+x^2}$

**mr fantastic** · January 29th 2011, 08:07 PM

Originally Posted by HNCMATHS

Hi everyone, which way should I approach this one....

Differentiate with respect to x

$Y = 3 arccot 2$

I suppose you mean y = 3 arcot(2t) ....? (Which is exactly the point skeeter made in post #2 that you seem not to have grasped).

**Archie Meade** · January 30th 2011, 02:04 PM

Originally Posted by HNCMATHS

Hi everyone, which way should I approach this one....

Differentiate with respect to x

$Y = 3 arccot (2x)$

$arccot(x)=cot^{-1}(x)$

$\displaystyle\frac{d}{dx}cot^{-1}(x)=-\frac{1}{1+x^2}$

$\displaystyle\frac{d}{dx}cot^{-1}(u)=-\frac{1}{1+u^2}\;\frac{du}{dx}$ from the chain rule

which gives

$\displaystyle\frac{d}{dx}cot^{-1}(2x)=?$

Using the notation "t"

$\displaystyle\frac{d}{dt}cot^{-1}(2t)=-\frac{1}{1+(2t)^2}\;\frac{d}{dt}(2t)$

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