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Linear functions have a constant slope, but what is the rate of change for a non-linear function as we shift along its domain?
(SB) Let \((x_0 ,f(x_0 ))\) be a point on te graph of \(y=f(x)\).
The derivative of \(f\) at \(x_0\) is the slope of the tangent line to the graph of \(f\) at \((x_0 ,f(x_0 ))\).
There are some common ways of denoting derivative of funtion \(f\) at \(x_0\):
\(\displaystyle f^{\prime } (x_0 )\)
\(\displaystyle \frac{df}{dx}(x_0 )\)
\(\displaystyle \frac{dy}{dx}(x_0 )\)
\(f_x (x_0 )\): this is popular in economics
We write this analyticaly as:
\[f^{\prime } (x_0 )=\lim_{h\to 0} \frac{f(x_0 +h)-f(x_0 )}{h}\]
If this limit exists, then the function \(f\) is differentiable at \(x_0\).
We will use this formula to derive first order taylor approximation. And this will also appear when we derive the formula for point elasticity.
given constant \(k\),:
\(\displaystyle f(x)=a\cdot x\)
\(\displaystyle f^{\prime } (x_0 )=a\)
syms x a
f(x, a) = a*xf(x, a) =
1em \(\displaystyle a\,x\)
dfk = diff(f,x)dfk(x, a) =
1em \(\displaystyle a\)
(SB) For any positive integer \(k\) (or real number \(k\)), the derivative of \(f(x)=x^k\) at \(x_0\) is:
\(\displaystyle f(x)=x^k\)
\(\displaystyle f^{\prime } (x_0 )=k\cdot x_0^{k-1}\)
syms x a k
f(x, a, k) = a*x^kf(x, a, k) =
1em \(\displaystyle a\,x^k\)
dfk = diff(f,x)dfk(x, a, k) =
1em \(\displaystyle a\,k\,x^{k-1}\)
\(\displaystyle f(x)=p(q(x))\)
\(\displaystyle f^{\prime } (x_0 )=p^{\prime } (q(x_0 ))\cdot q^{\prime } (x_0 )\)
syms x a k
f(x, a, k) = (a*x)^kf(x, a, k) =
1em \(\displaystyle {{\left(a\,x\right)}}^k\)
dfk = diff(f,x)dfk(x, a, k) =
1em \(\displaystyle a\,k\,{{\left(a\,x\right)}}^{k-1}\)
Given functions \(p\) and \(q\) that are differentiable at \(x\), then:
\(\displaystyle f(x)=p(x)+q(x)\)
\(\displaystyle f^{\prime } (x)=p^{\prime } (x)+q^{\prime } (x)\)
syms x a b c d
f(x, a, b, c, d) = a*x^b + c*x^df(x, a, b, c, d) =
1em \(\displaystyle a\,x^b +c\,x^d\)
dfk = diff(f,x)dfk(x, a, b, c, d) =
1em \(\displaystyle a\,b\,x^{b-1} +c\,d\,x^{d-1}\)
Given functions \(p\) and \(q\) that are differentiable at \(x\), then:
\(\displaystyle f(x)=p(x)\cdot q(x)\)
\(\displaystyle f^{\prime } (x)=p^{\prime } (x)\cdot q(x)+p(x)\cdot q^{\prime } (x)\)
syms x a b c d
f(x, a, b, c) = (a*x^b)*(c*x^d)f(x, a, b, c) =
1em \(\displaystyle a\,c\,x^b \,x^d\)
dfk = diff(f,x)dfk(x, a, b, c) =
1em \(\displaystyle a\,b\,c\,x^d \,x^{b-1} +a\,c\,d\,x^b \,x^{d-1}\)
Given functions \(p\) and \(q\) that are differentiable at \(x\), then:
\(\displaystyle f(x)=\frac{p(x)}{q(x)}\)
\(\displaystyle f^{\prime } (x)=\frac{p^{\prime } (x)\cdot q(x)-p(x)\cdot q^{\prime } (x)}{(q(x))^2 }\)
Note that the quotient rule is based on the product rule, because:
So you can derive the quotient rule formula based on the product rule where the first term is \(p(x)\) and the second term is \(\frac{1}{q(x)}\).
syms x a b c d
f(x, a, b, c) = (a*x^b)/(c*x^d)f(x, a, b, c) =
1em \(\displaystyle \frac{a\,x^b }{c\,x^d }\)
dfk = diff(f,x)dfk(x, a, b, c) =
1em \(\displaystyle \frac{a\,b\,x^{b-1} }{c\,x^d }-\frac{a\,d\,x^b }{c\,x^{d+1} }\)
We use exponential functions in economnics a lot:
\(\displaystyle f(x)=\exp (a\cdot x)\)
\(\displaystyle f^{\prime } (x)=a\cdot \exp (a\cdot x)\)
syms x a
f(x, a) = exp(a*x)f(x, a) =
1em \(\displaystyle {\mathrm{e}}^{a\,x}\)
dfk = diff(f,x)dfk(x, a) =
1em \(\displaystyle a\,{\mathrm{e}}^{a\,x}\)
This is a special case of any power function
\(\displaystyle f(x)=c^{a\cdot x}\)
\(\displaystyle f^{\prime } (x)=a\cdot (\log c)\cdot c^{a\cdot x}\)
note that \(log(exp(c))=c\)
syms x a c
f(x, a, c) = c^(a*x)f(x, a, c) =
1em \(\displaystyle c^{a\,x}\)
dfk = diff(f,x)dfk(x, a, c) =
1em \(\displaystyle a\,c^{a\,x} \,\log \left(c\right)\)
We use Log functions in economnics a lot:
\(\displaystyle f(x)=\log (a\cdot x)\)
\(\displaystyle f^{\prime } (x)=\frac{1}{x}\)
note that the c cancels out.
syms x a
f(x, a) = log(a*x)f(x, a) =
1em \(\displaystyle \log \left(a\,x\right)\)
dfk = diff(f,x)dfk(x, a) =
1em \(\displaystyle \frac{1}{x}\)