Differentiation Rules
- Page ID
- 217585
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Differentiation Rules
Constant Rule and Power Rule We have seen the following derivatives:
This leads us the guess the following theorem.
\( (x + h)^n = x^n + nhx^{n-1} + _nC_2h^2x^{n-2} + ... + nh^{n-1}x +h^n x^n \) \( = x^n + nhx^{n-1} + h^2(\text{Stuff}) \) For our purposes, stuff is some polynomial in \(h\) and \(k\) that we do not care about. We have \( \displaystyle \lim_{h \to 0} \frac{(x + h)^n - x^n}{h} = \lim_{h \to 0} \frac{x^n + nhx^{n-1} + h^2(\text{Stuff})}{h} \) \( \displaystyle = \lim_{h \to 0} \frac{nhx^{n-1} + h^2(\text{Stuff})}{h} = \lim_{h \to 0} \frac{h[nx^{n-1} + h(\text{Stuff})]}{h} \) \( \displaystyle = \lim_{h \to 0} nx^{n-1} + h(\text{Stuff}) = nx^{n-1}\) Sum Difference and Constant Multiple Rules
Theorem: If f and g are differentiable then A) \( [f + g]' = f ' + g'\) B) \( [f - g]' = f ' - g' \) C) \( [cf]' = c(f ') \)
Proof of C) \( \displaystyle = \lim_{h \to 0} \frac{cf(x + h) - cf(x)}{h}\) \( \displaystyle = \lim_{h \to 0} \frac{c[f(x + h) - f(x)]}{h}\) \( \displaystyle = c \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\) \( \displaystyle = c f'(x)\)
Example We know that if \(f(x) = x^2 \) then \(f'(x) = 2x \) and that if \( f(x) = x \) then \(f'(x) = 1\) so that if \(h(x) = 4x^2 -3x\) then \( h'(x) = 4(2x) - 3(1) = 8x - 1\)
More Applications Example
Solution
Example: Solution: Example: Solution: \( x = 4 \text{ or } x = -2 \)
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