Derivatives in use

Implicit differentiation

Example 1

\(xy=5\)
[Note that this corresponds to the function \(y=f\left(x\right)=5x^{-1}\)]
Rewrite this as
\(xf\left(x\right)=5,\ \forall\ x\ne0\)
We take the derivative wrt to x of both sides
\(\frac{d}{dx}\left(xf\left(x\right)\right)=\frac{d}{dx}5\)
Using the product rule yields
\(f\left(x\right)+xf'\left(x\right)=0\) 
Solving for f'(x)
\(f'\left(x\right)=-\frac{f\left(x\right)}{x}\)

Example 2

\(y^3+3x^2y=13\)
We must use the chain rule and the product rule
\(3y^2y'+6xy+3xy'=0\)
Once again, we solve for y'
\(y'=-2\frac{xy}{y^2+x^2}\)

How to implicitly differentiate

  1. Differentiate each side of the implicit function wrt x, while considering y as a function of x
  2. Solve the resulting equation for y'

Example 3

\(x^2y^3+\left(y+1\right)e^{-x}=x+2\)
Rewrite
\(x^2f\left(x\right)^3+\left(f\left(x\right)+1\right)e^{-x}=x+2\)
We can now differentiate both sides
\(2xf\left(x\right)^{3}+3x^{2}f\left(x\right)^{2}f^{'}\left(x\right)+f^{'}\left(x\right)e^{-x}-\left(f\left(x\right)+1\right)e^{-x}=1\)
Solving for f'(x)
\(f^{'}\left(x\right)=\frac{1-2xf\left(x\right)^{3}+\left[f\left(x\right)+1\right]e^{-x}}{3x^{2}f\left(x\right)^{2}+e^{-x}}\)
Use the same exact method to find the second implicit derivative

Economic examples

Effect of a per unit tax

\(D:\ Q_d=a-b(P+T)\)
\(S:\ Q_s=\alpha+\beta(P)\)
In equilibrium
\(a-b(P+T)=\alpha+\beta P\)
Taking the implicit derivative wrt T
\(-b(\frac{dP}{dT}+1)=\alpha+\beta(\frac{dP}{dT})\)
Solving for \(\frac{dP}{dT}\)
\(\frac{dP}{dT}=-\frac{b}{\beta+b}\)
Thus, taxes imposed on the consumer side reduce the equilibrium price

Differentiating the inverse function

Example 1

\(f(x)=ax+b\)
\(g(x)=\frac{1}{a}x-\frac{b}{a}\)
Where g is the inverse of f
\(f'(x)=a\)
\(g^{'}(x)=\frac{1}{a}\)
In general, when g(x) is the inverse of f(x)
\(g(f(x))=x,\ \forall x\ in\ I\)
Implicitly differentiating wrt x we get
\(\)\(g^{'}(f(x))f^{'}(x)=1\)
Solving for g'
\(g^{'}(f(x))=\frac{1}{f^{'}(x)}\)
Thus, the first derivative of the inverse of f(x) is equal to one over the first derivative of f(x)

Example 2

\(f(x)=x^{5}+3x^{3}+6x-3\)
\(f^{'}(x)=5x^{4}+9x^{2}+6\)
Thus
\(g^{'}(x)=\frac{1}{5x^{4}+9x^{2}+6}\)

Linear approximations

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Polynomial approximations

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Taylor approximation

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Elasticities

\(Q=D(P)\)
\(\epsilon_P=\frac{\Delta Q}{Q}=\frac{D(P+\Delta P)-D(P)}{D(P)}\)
Multiply both sides by \(\frac{P}{\Delta P}\)
\(\frac{\Delta Q}{Q}\frac{P}{\Delta P}=\frac{D(P+\Delta P)-D(P)}{D(P)}\frac{P}{\Delta P}=\frac{D(P+\Delta P)-D(P)}{\Delta P}\frac{P}{D(P)}\)

Example 1

Find \(\epsilon_x\) for \(f(x)=Ax^b\)
\(\epsilon_x=\frac{d f(x)}{d x}\frac{x}{f(x)}=f^{'}(x)\frac{x}{f(x)}\)
Finding f'(x) and substituting in f'(x) and f(x) yields
\(\epsilon_x=bAx^{b-1}\frac{x}{Ax^b}=b\)

Example 2

\(D(P)=8000P^{-1.5}\)
\(\epsilon_P=-12000P^{-2.5}\frac{P}{8000P^{-1.5}}=-1.5\)

Example 3

Take the following revenue function
\(R(P)=P\cdot Q(P)\)
Find the elasticity of revenue wrt price
\(\epsilon_R=R^{'}(P)\frac{P}{R(P)}\)
\(R^{'}(P)=\frac{d (P\cdot Q(P))}{dP}=Q(P)+PQ^{'}(P)\)
Thus, substituting in yields
\(\epsilon_R=\frac{Q(P)}{Q(P)}+Q^{'}(P)\frac{P}{Q(P)}\)
But \(Q^{'}(P)\frac{P}{Q(P)}\) is just \(\epsilon_P\)
Thus,
\(\epsilon_R=1+\epsilon_P\)

L'Hopital's Rule

\(\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{f^{'}(x)}{g^{'}(x)}\)
\(\lim_{x\to a}\frac{e^{x}-1}{x}=\frac{e^{x}}{1}=1\)

Example 1

\(\lim_{\lambda\to 0}\frac{x^{\lambda}-y^{\lambda}}{\lambda}=\frac{x^{lambda}\ln x-y^{\lambda}\ln y}{1}=\frac{\ln x-\ln y}{1}=\ln x-\ln y=\ln\frac{x}{y}\)

The constant elasticity of substitution (CES)

\(F(K,\ L)=A(aK^{-\sigma}+(1-a)L^{-\sigma})^{-1/\sigma}\)
Using l'Hopital's rule we can show that, as \(\sigma\) converges to zero, the function converges to the Cobb-Douglas functional form