Using differentiation from first principles, which expression represents the derivative \f… — JAMB UTME

Using differentiation from first principles, which expression represents the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the function $f(x) = \frac{2}{x}$ , where $x \neq 0$ ?

$-\frac{2}{x^2}$ Cevap
B
$\frac{2}{x^2}$
C
$-\frac{2}{x}$
D
$-\frac{1}{x^2}$

Cevap

-\frac{2}{x^2}

The option stating

-\frac{2}{x^2}

is correct because applying the first principles limit formula

\lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}

gives

\lim_{h \to 0} \frac{-2h}{h x(x+h)} = \lim_{h \to 0} \frac{-2}{x(x+h)} = -\frac{2}{x^2}

Adım Adım Çözüm

Set up the difference quotient definition of derivative from first principles.

\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

First principles definition states that the derivative is the limit of the rate of change as

h

approaches zero.

Substitute

f(x) = \frac{2}{x}

and

f(x+h) = \frac{2}{x+h}

into the formula.

\frac{f(x+h) - f(x)}{h} = \frac{\frac{2}{x+h} - \frac{2}{x}}{h}

Replacing the function expressions yields the explicit algebraic fraction.

Combine the fractions in the numerator over a common denominator.

\frac{2x - 2(x+h)}{x(x+h) \cdot h} = \frac{2x - 2x - 2h}{h x(x+h)} = \frac{-2h}{h x(x+h)}

Simplifying the numerator yields a factor of

h

that can be cancelled with the denominator.

Cancel the common factor

h

and evaluate the limit as

h \to 0

\lim_{h \to 0} \frac{-2}{x(x+h)} = \frac{-2}{x(x+0)} = -\frac{2}{x^2}

Dividing out

h

allows direct substitution of

h = 0

without producing an indeterminate form.

Anahtar Kavram

Differentiation from First Principles for Reciprocal Functions

Tahmini Süre:1m 30s

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