Question

How do I find the derivative of `f(x)=sqrt(x)` using first principles?

Answer 1

Definition

`f'(x)=lim_{h to 0}{f(x+h)-f(x)}/h`

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By Definition,

`f'(x)=lim_{h to 0}{sqrt{x+h}-sqrt{x}}/h`

by multiplying the numerator and the denominator by `sqrt{x+h}+sqrt{x}`,

`=lim_{h to 0}{sqrt{x+h}-sqrt{x}}/hcdot{sqrt{x+h}+sqrt{x}}/{sqrt{x+h}+sqrt{x}}`

`=lim_{h to 0}{x+h-x}/{h(sqrt{x+h}+sqrt{x})}`

by cancelling out `x`'s and `h`'s,

`=lim_{h to 0}1/{sqrt{x+h}+sqrt{x}}=1/{sqrt{x+0}+sqrt{x}}=1/{2sqrt{x}}`

Hence, `f'(x)=1/{2sqrt{x}}`.

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I hope that this was helpful.