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Thread: Implicit Differentiation Problem

  1. #1
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    Implicit Differentiation Problem

    Implicit Differentiation Problem-capture.png

    I was wondering if I am doing this right because i have no clue if i am doing it right!

    So for f(x) .. i calculated its derivative:

    arcsin(x/a)
    same as y= arcsin(x/a)
    sam as x/a= sin(y)

    implicit differentiation , i get

    1/a = cosy *dy/dx
    dy/dx = 1/acosy (is this the most simplified form?

    For g(x) . the derivative is

    g(x) = a*arctan(x/a)
    same as y= a*arctan(x/a)
    same as siny = a *(x/a)
    same as siny = x
    derivative through implicit differentiation, i get

    dy/dx[siny=x]
    dy/dx *cosy = 1
    dy/dx = 1/cosy

    Suppose i found the derivatives for f(x) and g(x) in its most simplest form.. how do i go about finding a? Do i set them equal to each other and solve for a?


    Thanks ;/!
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  2. #2
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    Re: Implicit Differentiation Problem

    Quote Originally Posted by lc99 View Post
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    I was wondering if I am doing this right because i have no clue if i am doing it right!

    So for f(x) .. i calculated its derivative:

    arcsin(x/a)
    same as y= arcsin(x/a)
    sam as x/a= sin(y)

    implicit differentiation , i get

    1/a = cosy *dy/dx
    dy/dx = 1/acosy (is this the most simplified form?

    For g(x) . the derivative is

    g(x) = a*arctan(x/a)
    same as y= a*arctan(x/a)
    same as siny = a *(x/a)
    same as siny = x
    derivative through implicit differentiation, i get

    dy/dx[siny=x]
    dy/dx *cosy = 1
    dy/dx = 1/cosy

    Suppose i found the derivatives for f(x) and g(x) in its most simplest form.. how do i go about finding a? Do i set them equal to each other and solve for a?


    Thanks ;/!
    Remember that you replaced $f(x)$ with $y$ and $g(x)$ with $y$, but $f(x) \neq g(x)$. So, a better way of writing it would be:

    $\sin f(x) = \dfrac{x}{a}$

    $\tan \left[ \dfrac{g(x)}{a}\right] = \dfrac{x}{a}$

    Then, you get $\dfrac{d f(x)}{dx} = \dfrac{1}{a \cos f(x)}$.

    For the second one, you were way off:

    $\dfrac{1}{a} \sec^2 \left[ \dfrac{g(x)}{a}\right] \dfrac{d g(x)}{dx} = \dfrac{1}{a}$ implies $\dfrac{d g(x)}{dx} = \cos^2 \left[ \dfrac{g(x)}{a} \right]$

    Now, setting them equal becomes more difficult. You want to get it back in terms of $x$.

    $\sin f(x) = \dfrac{x}{a} = \dfrac{\text{opp}}{\text{hyp}}$. You have a right-triangle with a side of length $x$ and hypotenuse of length $a$. Therefore the other side is $\sqrt{a^2-x^2}$ by the Pythagorean Theorem.
    You want $\cos f(x) = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{\sqrt{a^2-x^2}}{a}$

    So $\dfrac{d f(x)}{dx} = \dfrac{1}{\sqrt{a^2-x^2}}$

    Next, you have $\tan \left[ \dfrac{g(x)}{a} \right] = \dfrac{x}{a} = \dfrac{\text{opp}}{\text{adj}}$. You have a right-triangle with sides of length $x$ and $a$, so the hypotenuse is of length $\sqrt{a^2+x^2}$. So, $\cos \left[ \dfrac{g(x)}{a} \right] = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{a}{\sqrt{a^2+x^2}}$

    Thus, you have:

    $\dfrac{d g(x)}{dx} = \dfrac{a^2}{a^2+x^2}$

    Now, you want the derivatives at $x=1$ to be equal:

    $\dfrac{1}{\sqrt{a^2-1}} = \dfrac{a^2}{a^2+1}$

    Can you solve it from here?
    Last edited by SlipEternal; Oct 11th 2017 at 11:01 AM.
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