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Miscellaneous Examples · Example 19

Q.Find the derivative of ff from the first principle, where ff is given by

(i) f(x)=2x+3x−2f(x) = \dfrac{2x + 3}{x - 2}
(ii) f(x)=x+1xf(x) = x + \dfrac{1}{x}
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✓ Free question

Use the definition f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} to find derivatives from first principles. For (i) f(x)=2x+3x−2f(x) = \frac{2x+3}{x-2}, the derivative is −7(x−2)2\boxed{-\frac{7}{(x-2)^2}}; for (ii) f(x)=x+1xf(x) = x + \frac{1}{x}, the derivative is 1−1x2\boxed{1 - \frac{1}{x^2}}.

The derivative from first principles captures the instantaneous rate of change by examining what happens to the difference quotient as the interval shrinks to zero. Instead of applying ready-made rules, we return to the fundamental definition: the derivative at xx is the limit of the average rate of change over an interval [x,x+h][x, x+h] as hh approaches zero.

This means we compute f(x+h)f(x+h), subtract f(x)f(x), divide by hh, and then take the limit. The algebra often looks messy at first, but simplification always reveals the derivative.


(i) f(x)=2x+3x−2f(x) = \dfrac{2x + 3}{x - 2}

  1. Write the definition

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  1. Compute f(x+h)f(x+h) Replace xx with x+hx+h:

f(x+h)=2(x+h)+3(x+h)−2=2x+2h+3x+h−2f(x+h) = \frac{2(x+h) + 3}{(x+h) - 2} = \frac{2x + 2h + 3}{x + h - 2}

  1. Form the difference f(x+h)−f(x)f(x+h) - f(x)

f(x+h)−f(x)=2x+2h+3x+h−2−2x+3x−2f(x+h) - f(x) = \frac{2x + 2h + 3}{x + h - 2} - \frac{2x + 3}{x - 2}

To subtract these fractions, find a common denominator (x+h−2)(x−2)(x+h-2)(x-2):

=(2x+2h+3)(x−2)−(2x+3)(x+h−2)(x+h−2)(x−2)= \frac{(2x + 2h + 3)(x - 2) - (2x + 3)(x + h - 2)}{(x+h-2)(x-2)}

  1. Expand the numerators First term:

(2x+2h+3)(x−2)=2x2−4x+2hx−4h+3x−6=2x2−x+2hx−4h−6(2x + 2h + 3)(x - 2) = 2x^2 - 4x + 2hx - 4h + 3x - 6 = 2x^2 - x + 2hx - 4h - 6

Second term:

(2x+3)(x+h−2)=2x2+2xh−4x+3x+3h−6=2x2−x+2xh+3h−6(2x + 3)(x + h - 2) = 2x^2 + 2xh - 4x + 3x + 3h - 6 = 2x^2 - x + 2xh + 3h - 6

Subtract:

2x2−x+2hx−4h−6−(2x2−x+2xh+3h−6)=2hx−4h−2xh−3h=−7h2x^2 - x + 2hx - 4h - 6 - (2x^2 - x + 2xh + 3h - 6) = 2hx - 4h - 2xh - 3h = -7h

  1. Divide by hh

f(x+h)−f(x)h=−7hh(x+h−2)(x−2)=−7(x+h−2)(x−2)\frac{f(x+h) - f(x)}{h} = \frac{-7h}{h(x+h-2)(x-2)} = \frac{-7}{(x+h-2)(x-2)}

  1. Take the limit as h→0h \to 0

f′(x)=lim⁡h→0−7(x+h−2)(x−2)=−7(x−2)(x−2)=−7(x−2)2f'(x) = \lim_{h \to 0} \frac{-7}{(x+h-2)(x-2)} = \frac{-7}{(x-2)(x-2)} = -\frac{7}{(x-2)^2}

Tip

When subtracting rational functions in first-principles problems, the hh in the numerator always cancels with the hh in the denominator after simplification — that's what makes the limit exist.


(ii) f(x)=x+1xf(x) = x + \dfrac{1}{x}

  1. Write the definition

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  1. Compute f(x+h)f(x+h)

f(x+h)=(x+h)+1x+hf(x+h) = (x+h) + \frac{1}{x+h}

  1. Form the difference f(x+h)−f(x)f(x+h) - f(x)

f(x+h)−f(x)=(x+h+1x+h)−(x+1x)=h+1x+h−1xf(x+h) - f(x) = \left(x+h + \frac{1}{x+h}\right) - \left(x + \frac{1}{x}\right) = h + \frac{1}{x+h} - \frac{1}{x}

  1. Simplify the fraction difference

1x+h−1x=x−(x+h)x(x+h)=−hx(x+h)\frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}

So:

f(x+h)−f(x)=h−hx(x+h)=h(1−1x(x+h))f(x+h) - f(x) = h - \frac{h}{x(x+h)} = h\left(1 - \frac{1}{x(x+h)}\right)

  1. Divide by hh

f(x+h)−f(x)h=1−1x(x+h)\frac{f(x+h) - f(x)}{h} = 1 - \frac{1}{x(x+h)}

  1. Take the limit as h→0h \to 0

f′(x)=lim⁡h→0(1−1x(x+h))=1−1x⋅x=1−1x2f'(x) = \lim_{h \to 0} \left(1 - \frac{1}{x(x+h)}\right) = 1 - \frac{1}{x \cdot x} = 1 - \frac{1}{x^2}

Watch out

A common mistake is forgetting to combine terms properly before dividing by hh. Always factor out hh from the entire numerator to ensure clean cancellation.


✓Final answer

For (i) f(x)=2x+3x−2f(x) = \frac{2x+3}{x-2}, the derivative is f′(x)=−7(x−2)2\boxed{f'(x) = -\frac{7}{(x-2)^2}}. For (ii) f(x)=x+1xf(x) = x + \frac{1}{x}, the derivative is f′(x)=1−1x2\boxed{f'(x) = 1 - \frac{1}{x^2}}.

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