How do you find the derivative of f(x)=12x−1 by first principles?

f(x)=(2x-1)You want the derivative of f to the power −12Chain rule−122(2x−1)−12−1The factor 2 comes from the derivative of f itself

f(x)=12x−1f′(a)=limh→0f(a+h)−f(a)h=limh→012a+2h−1−12a−1h=limh→02a−1−2a+2h−1h2a+2h−12a−1=limh→0(2a−1−2a+2h−1)(2a−1+2a+2h−1)h2a+2h−12a−1(2a−1+2a+2h−1)=limh→0(2a−1)−(2a+2h−1)h2a+2h−12a−1(2a−1+2a+2h−1)=limh→0(−2h}{h2a+2h−12a−1(2a−1+2a+2h−1)}=limh→0(−2}{h2a+2h−12a−1(2a−1+2a+2h−1)}=−22a−12a−1(2a−1+2a−1)=−1(2a−1)32=−(2a−1)−32So:f′(x)=−(2x−1)−32

f(x)=(2x-1)You want the derivative of f to the power −12Chain rule−122(2x−1)−12−1The factor 2 comes from the derivative of f itself

f(x)=12x−1f′(a)=limh→0f(a+h)−f(a)h=limh→012a+2h−1−12a−1h=limh→02a−1−2a+2h−1h2a+2h−12a−1=limh→0(2a−1−2a+2h−1)(2a−1+2a+2h−1)h2a+2h−12a−1(2a−1+2a+2h−1)=limh→0(2a−1)−(2a+2h−1)h2a+2h−12a−1(2a−1+2a+2h−1)=limh→0(−2h}{h2a+2h−12a−1(2a−1+2a+2h−1)}=limh→0(−2}{h2a+2h−12a−1(2a−1+2a+2h−1)}=−22a−12a−1(2a−1+2a−1)=−1(2a−1)32=−(2a−1)−32So:f′(x)=−(2x−1)−32

Question: "How do you find the derivative of f(x)=12x−1..."

Bryson Whitney

Answered question

2022-04-14

How do you find the derivative of

f (x) = \frac{1}{\sqrt{2 x - 1}}

by first principles?

Answer & Explanation

maggionmoo

Beginner2022-04-15Added 16 answers

f(x)=(2x-1)
You want the derivative of f to the power

- \frac{1}{2}

Chain rule

- \frac{1}{2} 2 {(2 x - 1)}^{- \frac{1}{2} - 1}

The factor 2 comes from the derivative of f itself

unduncjineei5r3

Beginner2022-04-16Added 19 answers

f (x) = \frac{1}{\sqrt{2 x - 1}}

f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h}

= lim_{h \to 0} \frac{\frac{1}{\sqrt{2 a + 2 h - 1}} - \frac{1}{\sqrt{2 a - 1}}}{h}

= lim_{h \to 0} \frac{\sqrt{2 a - 1} - \sqrt{2 a + 2 h - 1}}{h \sqrt{2 a + 2 h - 1} \sqrt{2 a - 1}}

= lim_{h \to 0} \frac{(\sqrt{2 a - 1} - \sqrt{2 a + 2 h - 1}) (\sqrt{2 a - 1} + \sqrt{2 a + 2 h - 1})}{h \sqrt{2 a + 2 h - 1} \sqrt{2 a - 1} (\sqrt{2 a - 1} + \sqrt{2 a + 2 h - 1})}

= lim_{h \to 0} \frac{(2 a - 1) - (2 a + 2 h - 1)}{h \sqrt{2 a + 2 h - 1} \sqrt{2 a - 1} (\sqrt{2 a - 1} + \sqrt{2 a + 2 h - 1})}

= lim_{h \to 0} \frac{(- 2 h} {h \sqrt{2 a + 2 h - 1} \sqrt{2 a - 1} (\sqrt{2 a - 1} + \sqrt{2 a + 2 h - 1})}}{}

= lim_{h \to 0} \frac{(- 2} {h \sqrt{2 a + 2 h - 1} \sqrt{2 a - 1} (\sqrt{2 a - 1} + \sqrt{2 a + 2 h - 1})}}{}

= \frac{- 2}{\sqrt{2 a - 1} \sqrt{2 a - 1} (\sqrt{2 a - 1} + \sqrt{2 a - 1})}

= \frac{- 1}{{(2 a - 1)}^{\frac{3}{2}}}

= - {(2 a - 1)}^{- \frac{3}{2}}

So:

f^{'} (x) = - {(2 x - 1)}^{- \frac{3}{2}}

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