Direct solution via the Context Panel

$x^2\+y^2\=9$$\stackrel{\text{implicit differentiation}}{\to }$$-\frac{x}{y}$

Solution by task template

Tools≻Tasks≻Browse:

Calculus - Differential≻Derivatives≻Implicit Differentiation≻$f\left(x\,y\right)\=g\left(x\,y\right)$ 

$\frac{d}{\mathrm{dx}}\left(x^2\+y{\left(x\right)}^2-9\right)$ 

embedded. The details of the differentiation can be explored via this tutor. Table 2.5.1(a) lists and annotates the steps as per the tutor.

Interactively implemented stepwise solution

$x^2\+y^2\=9$

$x^2\+y^2\=9$

$\stackrel{\text{evaluate at point}}{\to }$

$x^2\+{y\left(x\right)}^2\=9$

$\stackrel{\text{differentiate w.r.t. x}}{\to }$

$2x\+2y\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\=0$

$\stackrel{\text{isolate for diff(y(x),x)}}{\to }$

$\frac{\ⅆ}{\ⅆx}y\left(x\right)\=-\frac{x}{y\left(x\right)}$

Solution via the implicitdiff command

$\mathrm{implicitdiff}\left(x^2\+y^2\=9\,y\,x\right)$ = $-\frac{x}{y}$$$

Stepwise solution via the ImplicitDiffSolution command

$\mathrm{Student}:-\mathrm{Calculus1}:-\mathrm{ImplicitDiffSolution}\left(x^2\+y^2\=9\,y\,x\right)$

$\begin{array}{lll}& & \text{Implicit Differentiation Steps}\\ & & x^2+y^2=9\\ \text{•}& & \text{Rewrite}y\text{as a function}y\left(x\right)\text{:}\\ & & x^2+{y\left(x\right)}^2=9\\ \text{•}& & \text{Differentiate the left side}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2+{y\left(x\right)}^2\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(f\left(x\right)+g\left(x\right)\right)=\frac{\ⅆ}{\ⅆx}f\left(x\right)+\frac{\ⅆ}{\ⅆx}g\left(x\right)\\ & & f\left(x\right)=x^2\\ & & g\left(x\right)={y\left(x\right)}^2\\ & & \text{This gives:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)+\frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^2\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(x^2\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\partial }{\partial x}\left(x^n\right)=n{x}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=2\cdot x^1\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=2\cdot x\\ & & \text{We can rewrite the derivative as:}\\ & & \left(2x\right)+\frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^2\right)\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{chain}}\text{rule to the term}{y\left(x\right)}^2\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{chain}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)=\mathrm{f\text{'}}\left(g\left(x\right)\right)\left(\frac{\ⅆ}{\ⅆx}g\left(x\right)\right)\\ & \text{◦}& \text{Outside function}\\ & & f\left(v\right)=v^2\\ & \text{◦}& \text{Inside function}\\ & & g\left(x\right)=y\left(x\right)\\ & \text{◦}& \text{Derivative of outside function}\\ & & \frac{\ⅆ}{\ⅆv}f\left(v\right)=2v\\ & \text{◦}& \text{Apply composition}\\ & & \mathrm{f\text{'}}\left(g\left(x\right)\right)=2y\left(x\right)\\ & \text{◦}& \text{Derivative of inside function}\\ & & \frac{\ⅆ}{\ⅆx}g\left(x\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & \text{◦}& \text{Put it all together}\\ & & \left(\frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)\right)\left(\frac{\ⅆ}{\ⅆx}g\left(x\right)\right)=2y\left(x\right)\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\\ & & \text{This gives:}\\ & & 2x+\left(2y\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\right)\\ \text{•}& & \text{The final result is}\\ & & 2x+2\cdot y\left(x\right)\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\\ \text{•}& & \text{Differentiate the right side}\\ & & \frac{\ⅆ}{\ⅆx}9\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{constant}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}9\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \frac{\ⅆC}{\ⅆx}=0\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}9=0\\ & & \text{We can now rewrite the derivative as:}\\ & & 0\\ \text{•}& & \text{Rewrite}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{as}\mathrm{y\text{'}}\text{and solve for}\mathrm{y\text{'}}\\ & & 2x+2\cdot y\cdot \mathrm{y\text{'}}=0\\ \text{•}& & \text{Subtract}2\cdot x\text{from both sides}\\ & & 2\cdot x+2\cdot y\cdot \mathrm{y\text{'}}-2\cdot x=0-2\cdot x\\ \text{•}& & \text{Simplify}\\ & & 2\cdot y\cdot \mathrm{y\text{'}}=\left(\mathrm{-2}\right)\cdot x\\ \text{•}& & \text{Divide both sides by}2\cdot y\\ & & \frac{\mathrm{y\text{'}}\cdot \left(2\cdot y\right)}{2\cdot y}=\frac{\left(\mathrm{-2}\right)\cdot x}{2\cdot y}\\ \text{•}& & \text{Simplify}\\ & & \mathrm{y\text{'}}=\frac{-2x}{2\cdot y}\\ \text{•}& & \text{Reduce fraction by gcd}\\ & & \mathrm{y\text{'}}=-\frac{x}{y}\end{array}$