Exercise 214Calculus › Single-variable calculus › Differentiation › Differentiable functionsHigh schoolAdded 2 Nov 2026

The derivative from the definition

#definition of the derivative#difference quotient#square root

Statement

Using only the definition $f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}h$ :

compute the derivative of $f(x)=\dfrac1x$ at $a\ne0$ ,
compute the derivative of $g(x)=\sqrt x$ at $a>0$ ,
explain what goes wrong for $g$ at $a=0$ .

Hint

For part 1, put the numerator over a common denominator. For part 2, multiply by the conjugate.

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Solution

1.
$\frac{\frac1{a+h}-\frac1a}h=\frac{a-(a+h)}{h\,a(a+h)}=\frac{-1}{a(a+h)}\xrightarrow[h\to0]{}-\frac1{a^2}.$

2.
$\frac{\sqrt{a+h}-\sqrt a}h=\frac{(a+h)-a}{h\left(\sqrt{a+h}+\sqrt a\right)}=\frac1{\sqrt{a+h}+\sqrt a}\xrightarrow[h\to0]{}\frac1{2\sqrt a}.$

3. At $a=0$ the difference quotient is $\dfrac{\sqrt h}h=\dfrac1{\sqrt h}$ for $h>0$ , which tends to $+\infty$ . There is no finite limit: $\sqrt x$ is not differentiable at $0$ . Its graph has a vertical tangent there.

The derivative from the definition

Statement

Solution

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