### Differentiation, derivatives and Taylor approximations: Differentiating implicit functions

### Implicit differentiation

Implicit differentiation We have already noted that quantities are often related to each other through an implicit relationship. For example, you can write a linear relationship between \(x\) and \(y\) as the equation \[a\,x+b\,y+c=0\] for certain constants \(a\), \(b\) and \(c\). If \(b\neq 0\), then you can isolate \(y\) \[y=-\frac{a}{b}x-\frac{c}{b}\] and then consider\(y\) as a function of \(x\). For the derivative of \(y\) as a function of \(x\) you get: \[\frac{\dd y}{\dd x}=-\frac{a}{b}\]

But we could also have determined this derivative by applying the differential operator \(\frac{\dd }{\dd x}\) directly to the implicit relationship between \(x\) and \(y\) and using calculation rules for differentiation: \[\begin{aligned}\frac{\dd }{\dd x}(a\,x+b\,y+c)&=\frac{\dd }{\dd x}(a\,x)+\frac{\dd }{\dd x}(b\,y)+ \frac{\dd }{\dd x}(c)\\&\phantom{abcpqrxyz}\blue{\text{sum rule for differentiation}}\\[0.25cm] &= a\frac{\dd x}{\dd x}+ b\frac{\dd y}{\dd x}+0&\\&\phantom{abcpqrxyz}\blue{\text{constant factor rule, derivative of a constant}}\\[0.25cm] &= a+b\frac{\dd y}{\dd x}\end{aligned}\] Because of \(a\,x+b\,y+c=0\), the previous result must also be equal to zero. The derivative can then be isolated in \(a+b\frac{\dd y}{\dd x}=0\) and you get the same result as before. The second approach is called the method of **implicit differentiation**.

We present two more examples of implicit differentiation.

Slope of a circle at a point We look at the circle with equation \(x^2+y^2=25\) and ask ourselves what is the slope of the tangent line at the point \((-3,4)\).

We consider \(y\) as a function of \(x\) in the given equation and apply the differential operator \(\frac{\dd }{\dd x}\) to the left- and right-hand side of the equation; applying the rules of differentiation gives: \[\begin{aligned}\frac{\dd }{\dd x}(x^2)+\frac{\dd }{\dd x}(y^2)&=\frac{\dd }{\dd x}(25)\\[0.25cm] 2x+2y\frac{\dd y }{\dd x}&=0\\[0.25cm] \frac{\dd y }{\dd x} &= -\frac{x}{y}\end{aligned}\] The slope in \((-3,4)\) is therefore \(-\frac{x}{y}\biggl|_{(-3,4)}=-\frac{-3}{4}=\frac{3}{4}\).

Tangent line of a curve at a point We look at the curve with equation \(x^3+xy+y^3=5\) and want to know the equation of the tangent line at the point \((2,-1)\).

Substitution of \(x=2\) and \(y=-1\) in the equation of the curve shows that \((2,-1)\) is indeed a point on the curve. We consider \(y\) as a function of \(x\) in the given equation and apply the differential operator \(\frac{\dd }{\dd x}\) left and right on the equation; application of the rules of differentiation gives: \[\begin{aligned}\frac{\dd }{\dd x}(x^3)+\frac{\dd }{\dd x}(xy)+\frac{\dd }{\dd x}(y^3)&=\frac{\dd }{\dd x}(5)\\[0.25cm] 3x^2+y+ x\frac{\dd y}{\dd x}+3y^2\frac{\dd y }{\dd x}&=0\\[0.25cm] 3x^2+y+ (x+3y^2)\frac{\dd y }{\dd x}&=0\end{aligned}\] Substitution of \(x=2\) and \(y=-1\) gives: \[\begin{aligned}3\times 2^2-1+\bigl(2+3\times(-1)^2\bigr)y'(2)&=0\\[0.25cm] 11+5y'(2)&=0\\[0.25cm]y'(2)&=-\frac{11}{5}\end{aligned}\] The equation of the tangent line at the point \((2,-1)\) is \[y=-\tfrac{11}{5}(x-2)-1\qquad\text{or}\qquad y=-\tfrac{11}{5}x+\tfrac{17}{5}\]