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

  1. #1
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    Exclamation Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

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

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

    F'(x)=( (-x^2 -6x +5)/(x^2 +5)^2 ) / ( 1 + ((x+3)/(x^2 +5))^2 ) ?
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    Re: Help! Find f'(x) if f(x)= arccot ((x+3)/ (x^2 +5))

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

    Quote Originally Posted by nancyaldape View Post
    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*}
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