# Thread: Differentiate with respect to x

1. ## Differentiate with respect to x

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

Differentiate with respect to x

$\displaystyle Y = 3 arccot 2$

2. Originally Posted by HNCMATHS
Hi everyone, which way should I approach this one....

Differentiate with respect to x

$\displaystyle Y = 3 arccot 2$
since $\displaystyle 3arccot(2)$ is a constant, the derivative is 0.

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

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

3. Originally Posted by skeeter
since $\displaystyle 3\arctan(2)$ is a constant, the derivative is 0.

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

$\displaystyle \dfrac{d}{dx} [k\arctan{u}] = \dfrac{k}{1 + u^2} \cdot \dfrac{du}{dx}$
Hi thanks for the reply, however the problem I am working with is definately 3arccot 2t

4. Originally Posted by HNCMATHS
Hi everyone, which way should I approach this one....

Differentiate with respect to x

$\displaystyle Y = 3 arccot 2$
Well, you can either just accept that

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

or

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

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

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

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

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

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

5. Originally Posted by HNCMATHS
Hi everyone, which way should I approach this one....

Differentiate with respect to x

$\displaystyle Y = 3 arccot 2$
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).

6. Originally Posted by HNCMATHS
Hi everyone, which way should I approach this one....

Differentiate with respect to x

$\displaystyle Y = 3 arccot (2x)$
$\displaystyle arccot(x)=cot^{-1}(x)$

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

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

which gives

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

Using the notation "t"

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