Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

**nancyaldape** · January 17th 2013, 07:08 PM

Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

**MarkFL** · January 17th 2013, 07:19 PM

Use $\frac{d}{du}\left(\cot^{-1}(u(x)) \right)=-\frac{1}{u^2(x)+1}\cdot\frac{du}{dx}$

where $u(x)=\frac{x+3}{x^2+5}$

What do you find?

**nancyaldape** · January 17th 2013, 07:38 PM

F'(x)=( (-x^2 -6x +5)/(x^2 +5)^2 ) / ( 1 + ((x+3)/(x^2 +5))^2 ) ?

**MarkFL** · January 17th 2013, 07:56 PM

It appears you did not apply the negative sign in front of the derivative of the inverse cotangent function. Other than that, your result is correct, you just need to simplify by multiplying the result by:

$1=\frac{(x^2+5)^2}{(x^2+5)^2}$

**Prove It** · January 17th 2013, 09:34 PM

Originally Posted by nancyaldape

Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

I find that finding derivatives of inverse trigonometric functions are easiest to find using Implicit Differentiation.

$\displaystyle \begin{align*} y &= \textrm{arccot}\,{\left( \frac{x+3}{x^2+5} \right)} \\ \cot{(y)} &= \frac{x+3}{x^2+5} \\ \frac{d}{dx}\left[ \cot{(y)} \right] &= \frac{d}{dx} \left( \frac{x+3}{x^2+5} \right) \\ \frac{d}{dy}\left[ \frac{\cos{(y)}}{\sin{(y)}} \right] \frac{dy}{dx} &= \frac{1 \left( x^2 + 5 \right) - 2x \left( x + 3 \right)}{\left( x^2 + 5 \right) ^2 } \\ \left[ \frac{-\sin^2{(y)} - \cos^2{(y)}}{\sin^2{(y)}} \right] \frac{dy}{dx} &= \frac{5 - 6x - x^2}{\left( x^2 + 5 \right)^2} \\ \left[ -\frac{1}{\sin^2{(y)}} \right] \frac{dy}{dx} &= \frac{5 - 6x - x^2}{\left( x^2 + 5 \right)^2 } \\ \frac{dy}{dx} &= -\sin^2{(y)} \left[ \frac{5 - 6x - x^2}{\left( x^2 + 5 \right)^2} \right] \\ \frac{dy}{dx} &= -\frac{1}{\csc^2{(y)}} \left[ \frac{5 - 6x - x^2}{\left( x^2 + 5 \right)^2} \right] \\ \frac{dy}{dx} &= -\frac{1}{1 + \cot^2{(y)}} \left[ \frac{5 - 6x - x^2}{ \left( x^2 + 5 \right)^2} \right] \end{align*}$

$\displaystyle \begin{align*} \frac{dy}{dx} &= -\frac{1}{1 + \left\{ \cot{ \left[ \textrm{arccot}\,{ \left( \frac{x+3}{x^2 + 5} \right) } \right] } \right\}^2 } \left[ \frac{5 - 6x - x^2}{\left( x^2 + 5 \right)^2} \right] \\ \frac{dy}{dx} &= -\frac{1}{ 1 + \left( \frac{x+3}{x^2 + 5} \right)^2 } \left[ \frac{5 - 6x - x^2}{\left( x^2 + 5 \right)^2 } \right] \\ \frac{dy}{dx} &= -\frac{1}{\frac{\left( x^2 + 5 \right)^2 + \left( x + 3 \right)^2}{\left( x^2 + 5 \right)^2}} \left[ \frac{ 5 - 6x - x^2 }{\left( x^2 + 5 \right)^2 } \right] \\ \frac{dy}{dx} &= -\frac{ \left( x^2 + 5 \right)^2}{ x^4 + 11x^2 + 6x + 34} \left[ \frac{5 - 6x - x^2}{\left( x^2 + 5 \right)^2} \right] \\ \frac{dy}{dx} &= \frac{ x^2 + 6x - 5 }{ x^4 + 11x^2 + 6x + 34 } \end{align*}$

Thread: Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

LinkBack

Thread Tools

Display

Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

Re: Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

Re: Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

Re: Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

Re: Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

Similar Math Help Forum Discussions

arccos+arccot

proof of arccot(x)

What is the derivative of arccot(e^2x)?

arccos(3/5)+arccot(1/7)

Help me find Arccot please!

Search Tags