Title: CR_Diff_07
Course: Demonstrating STACK (v4.4.7)

Implicit differentiation

Find \(\frac{d[value]}{dx}\) in terms of [value] when \(x\) and [value] are related by the following equation:

\[x = [value]\]
Feedback: \[x=[value]\] So, \[\frac{dx}{dx}=2[value]\frac{d[value]}{dx}-\frac{1}{[value]^2}\frac{d[value]}{dx} \] So, taking common factor \(\frac{d[value]}{dx}\), \[1=\left(2[value] -\frac{1}{[value]^2}\right)\frac{d[value]}{dx}\] Hence, \[\begin{array}{rcl}\displaystyle \frac{d[value]}{dx} &\displaystyle =&\displaystyle \frac{1}{2[value] -\frac{1}{[value]^2}} \\ &\displaystyle =&\displaystyle \frac{[value]^2}{2[value]^3+1} \end{array}\]

Find \(\frac{d[value]}{dx}\) in terms of [value] when \(x\) and [value] are related by the following equation:

\[x = [value]\]

Find \(\frac{dy}{dx}\) where \([value]=0\) using implicit differentiation.
Feedback: Differentiating implicitly we obtain: \[ -\frac{1}{2}[value]-\frac{1}{2}\frac{dy}{dx}[value] = 0. \] Re-arranging we obtain: \[ \frac{dy}{dx} = [value]. \]

Find \(\frac{dy}{dx}\) where \([value]=0\) using implicit differentiation.
Feedback: \[y^2-y+x^2+x=0\] So, \[2y\frac{dy}{dx}-\frac{dy}{dx}+\frac{d}{dx}[x^2]+\frac{dx}{dx}=0\] By taking common factor \(\frac{dy}{dx},\) \[\frac{dy}{dx}(2y-1)=-2x-1\] Hence, \[\frac{dy}{dx}=\frac{-2x-1}{2y-1}\]

Find \(\frac{dy}{dx}\) where \([value]=0\) using implicit differentiation.
Feedback: Differentiating implicity with respect to x we obtain: \[ 1 + \frac{dy}{dx}[value] - [value] - \frac{dy}{dx}[value] = 0. \] \[ \frac{dy}{dx}\left([value]\right) = [value]. \] \[\frac{dy}{dx} = [value]. \] Rationalising and factorising gives: \[\frac{dy}{dx} = [value].\]