16. Techniques of Differentiation
Dated: 30-10-2024
Derivatives of Constant Functions
\[\frac{d}{dx} (c) = 0\]
Proof
\[\lim_{h \rightarrow 0} \frac{f(h + x) - f(x)}{h}\]
Since we know that \(f(h + x) = f(x) = c\).
\[ =\lim_{h \rightarrow 0} \frac{0 - 0}{h}\]
\[= \lim_{h \rightarrow 0} \frac{0}{h}\]
\[ = \lim_{h \rightarrow 0} 0\]
\[ = 0\]
Power Rule
\[\frac{d}{dx}(x^n) = n \cdot x^{n - 1}\]
Proof
\[\frac{d}{dx}(x^n) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\]
\[= \lim_{h \rightarrow 0} \frac{(x + h)^n - x^n}{h}\]
Using binomial theorem1
\[= \lim_{h \rightarrow 0} \frac{\left[x^n + nx^{n-1}h + \ldots + h^n\right] - x^n}{h}\]
\[= \lim_{h \rightarrow 0} \frac{nx^{n-1}h + \ldots + h^n}{h}\]
Taking \(h\) as common factor, out
\[= \lim_{h \rightarrow 0} \frac{h \left(nx^{n-1} + \ldots + h^{n1}\right)}{h}\]
\[= \lim_{h \rightarrow 0} \left(nx^{n-1} + \ldots + h^{n - 1}\right)\]
Since all the terms except for first one include some factor of \(h\), most of the terms become null.
\[= nx^{n-1}\]
Sum or Difference of the Functions
\[\frac{d}{dx}\left(f(x) + g(x)\right) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))\]
Proof
\[\frac{d}{dx}\left(f(x) + g(x)\right) = \lim_{h \rightarrow 0} \frac{f(x+h) + g(x + h) - f(x) - g(x)}{h}\]
\[\frac{d}{dx}\left(f(x) + g(x)\right) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x) + g(x + h) - g(x)}{h}\]
\[\frac{d}{dx}\left(f(x) + g(x)\right) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} + \lim_{h \rightarrow 0} \frac{g(x + h) - g(x)}{h}\]
\[\frac{d}{dx}\left(f(x) + g(x)\right) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)\]
Product
\[\frac{d}{dx}\left(f(x) \cdot g(x)\right) = f(x) \cdot \frac{d}{dx} g(x) + g(x) \cdot \frac{d}{dx} f(x)\]
Proof
\[\frac{d}{dx}\left(f(x) \cdot g(x)\right) = \lim_{h \rightarrow 0} \frac{f(x + h) \cdot g(x + h) - f(x) \cdot g(x)}{h}\]
\[= \lim_{h \rightarrow 0} \frac{f(x + h) \cdot g(x + h) - f(x) \cdot g(x)}{h}\]
Adding and subtracting \(f(x + h) \cdot g(x)\)
\[= \lim_{h \rightarrow 0} \frac{f(x + h) \cdot g(x + h) - f(x + h) \cdot g(x) + f(x + h) \cdot g(x) - f(x) \cdot g(x)}{h}\]
\[= \lim_{h \rightarrow 0} \frac{f(x + h) \cdot \left( g(x + h) - g(x) \right) + g(x) \cdot \left(f(x + h) - f(x) \right)}{h}\]
\[= \lim_{h \rightarrow 0} \frac{f(x + h) \cdot \left( g(x + h) - g(x) \right)}{h} + \lim_{h \rightarrow 0}\frac{g(x) \cdot \left(f(x + h) - f(x) \right)}{h}\]
\[= f(x) \cdot \frac{d}{dx} g(x) + g(x) \cdot \frac{d}{dx} f(x)\]
Quotient
\[\frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x) \frac{d}{dx}[f(x)] - f(x) \frac{d}{dx}[g(x)]}{[g(x)]^2}
\]
Proof
Reciprocal Theorem
\[\frac{d}{dx} \left( \frac{1}{g(x)} \right) = -\frac{\frac{d}{dx} \left( g(x) \right)}{\left( g(x) \right)^2}
\]
References
Read more about notations and symbols.
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Read more about binomial theorem. ↩