Using differentiation from first principles, what is the derivative \frac{\mathrm{d}y}{\ma… — JAMB UTME

Using differentiation from first principles, what is the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the function $f(x) = 3x^2 - 2x$ ?

The derivative of

f(x) = 3x^2 - 2x

with respect to

x

6x - 2

Applying the first principles definition

\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

f(x) = 3x^2 - 2x

yields

\frac{3x^2 + 6xh + 3h^2 - 2x - 2h - (3x^2 - 2x)}{h} = \frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2

. As

h \to 0

, this simplifies directly to

6x - 2

Set up the definition of the derivative from first principles.

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

The definition of the derivative is the limit of the difference quotient as

h

approaches zero.

Substitute

f(x+h)

and

f(x)

into the difference quotient.

f(x+h) = 3(x+h)^2 - 2(x+h) = 3(x^2 + 2xh + h^2) - 2x - 2h = 3x^2 + 6xh + 3h^2 - 2x - 2h

Expand

(x+h)^2

completely using algebraic identity.

Subtract

f(x)

from

f(x+h)

to find the numerator.

f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h) - (3x^2 - 2x) = 6xh + 3h^2 - 2h

Cancel out identical terms

3x^2

and

-2x

Divide by

h

and evaluate the limit as

h \to 0

\frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2; \quad \lim_{h \to 0} (6x + 3h - 2) = 6x - 2

Factor out

h

to cancel the denominator, then set

h = 0

Differentiation from First Principles

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