- Derivative (examples)
Example 1
Consider "f"("x") = 5:
: f'(x)=lim_{h ightarrow 0} frac{f(x+h)-f(x)}{h} = lim_{h ightarrow 0} frac{f(x+h)-5}{h} = lim_{h ightarrow 0} frac{(5-5)}{h} = lim_{h ightarrow 0} frac{0}{h} = lim_{h ightarrow 0} 0 = 0
The derivative of a
constant function is zero.Example 2
Consider the graph of f(x)=2x-3. If the reader has an understanding of
algebra and theCartesian coordinate system , the reader should be able to independently determine that this line has a slope of 2 at every point. Using the above quotient (along with an understanding of the limit,secant , and tangent) one can determine the slope at (4,5)::egin{align}f'(4) &= lim_{h o 0}frac{f(4+h)-f(4)}{h} \ &= lim_{h o 0}frac{2(4+h)-3-(2cdot 4-3)}{h} \ &= lim_{h o 0}frac{8+2h-3-8+3}{h} \ &= lim_{h o 0}frac{2h}{h} \ &= 2end{align}
The derivative and slope are equivalent.
Example 3
Via differentiation, one can find the slope of a curve. Consider f(x)=x^2:
:
For any point "x", the slope of the function f(x)=x^2 is f'(x)=2x.
Example 4
Consider f(x) = sqrt{x} :
:
Example 5
The same as the previous example, but now we search the derivative of the derivative.
Consider f(x) = sqrt{x} ::egin{align}f"(x) &= lim_{h o 0} frac{f'(x+h)-f'(x)}{h} \ &= lim_{h o 0} frac{frac{1}{2 sqrt{x+h-frac{1}{2 sqrt{x}{h} \ &= lim_{h o 0} frac{1}{2h}left(frac{1}{sqrt{x+h-frac{1}{sqrt{x ight) \ &= lim_{h o 0} frac{1}{2h}left(frac{sqrt{x} - sqrt{x+h{sqrt{x}sqrt{x+h ight) \ &= lim_{h o 0} frac{1}{2h}left(frac{sqrt{x} - sqrt{x+h{sqrt{x}sqrt{x+h imes frac{sqrt{x} + sqrt{x+h{sqrt{x} + sqrt{x+h ight) \ &= lim_{h o 0} frac{1}{2h}left(frac{x - (x+h)}{xsqrt{x+h} + (x+h)sqrt{x} } ight) \ &= lim_{h o 0} frac{1}{2}left(frac{-1}{xsqrt{x+h} + (x+h)sqrt{x} } ight) \ &= frac{1}{2}left(frac{-1}{xsqrt{x} + xsqrt{x} } ight) \ &= -frac{1}{4 x sqrt{xend{align}
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