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Question: differentiate: (x-a/x-b) with first principle...

differentiate: (x-a/x-b) with first principle

Answer

f'(x) = \frac{a-b}{(x-b)^2}

Explanation

Solution

To differentiate the function f(x)=x−ax−bf(x) = \frac{x-a}{x-b} using the first principle, we use the definition of the derivative:

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

Step 1: Find f(x+h)f(x+h)

Substitute (x+h)(x+h) for xx in the function:

f(x+h)=(x+h)−a(x+h)−b=x+h−ax+h−bf(x+h) = \frac{(x+h)-a}{(x+h)-b} = \frac{x+h-a}{x+h-b}

Step 2: Calculate f(x+h)−f(x)f(x+h) - f(x)

Subtract f(x)f(x) from f(x+h)f(x+h):

f(x+h)−f(x)=x+h−ax+h−b−x−ax−bf(x+h) - f(x) = \frac{x+h-a}{x+h-b} - \frac{x-a}{x-b}

To combine these fractions, find a common denominator, which is (x+h−b)(x−b)(x+h-b)(x-b):

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

Expand the numerator:

Numerator =(x(x−b)+h(x−b)−a(x−b))−(x(x+h−b)−a(x+h−b)) = (x(x-b) + h(x-b) - a(x-b)) - (x(x+h-b) - a(x+h-b))

Numerator =(x2−xb+hx−hb−ax+ab)−(x2+xh−xb−ax−ah+ab) = (x^2 - xb + hx - hb - ax + ab) - (x^2 + xh - xb - ax - ah + ab)

Now, remove the parentheses and change the signs of the terms inside the second parenthesis:

Numerator =x2−xb+hx−hb−ax+ab−x2−xh+xb+ax+ah−ab = x^2 - xb + hx - hb - ax + ab - x^2 - xh + xb + ax + ah - ab

Group and cancel out terms:

Numerator =(x2−x2)+(−xb+xb)+(hx−xh)+(−ax+ax)+(ab−ab)−hb+ah = (x^2 - x^2) + (-xb + xb) + (hx - xh) + (-ax + ax) + (ab - ab) - hb + ah

Numerator =0+0+0+0−hb+ah = 0 + 0 + 0 + 0 - hb + ah

Numerator =ah−hb = ah - hb

Factor out hh from the numerator:

Numerator =h(a−b) = h(a-b)

So, f(x+h)−f(x)=h(a−b)(x+h−b)(x−b)f(x+h) - f(x) = \frac{h(a-b)}{(x+h-b)(x-b)}

Step 3: Divide by hh

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

Cancel out hh (since h≠0h \neq 0 for the limit process):

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

Step 4: Take the limit as h→0h \to 0

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

Substitute h=0h=0 into the expression:

f′(x)=a−b(x+0−b)(x−b)f'(x) = \frac{a-b}{(x+0-b)(x-b)}

f′(x)=a−b(x−b)(x−b)f'(x) = \frac{a-b}{(x-b)(x-b)}

f′(x)=a−b(x−b)2f'(x) = \frac{a-b}{(x-b)^2}

The derivative of x−ax−b\frac{x-a}{x-b} with respect to xx is a−b(x−b)2\frac{a-b}{(x-b)^2}.

Explanation of the solution:

The derivative of f(x)=x−ax−bf(x) = \frac{x-a}{x-b} is found using the first principle definition: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.

  1. Substitute x+hx+h into f(x)f(x) to get f(x+h)=x+h−ax+h−bf(x+h) = \frac{x+h-a}{x+h-b}.
  2. Calculate the difference f(x+h)−f(x)=x+h−ax+h−b−x−ax−bf(x+h) - f(x) = \frac{x+h-a}{x+h-b} - \frac{x-a}{x-b}.
  3. Combine the fractions by finding a common denominator (x+h−b)(x−b)(x+h-b)(x-b). The numerator simplifies to h(a−b)h(a-b) after expanding and canceling terms.
  4. Divide the expression by hh, which cancels out the hh in the numerator. This leaves a−b(x+h−b)(x−b)\frac{a-b}{(x+h-b)(x-b)}.
  5. Take the limit as h→0h \to 0. Substituting h=0h=0 yields the final derivative: a−b(x−b)2\frac{a-b}{(x-b)^2}.

Answer:

The derivative of (x−a/x−b)(x-a/x-b) with respect to xx using the first principle is:

f′(x)=a−b(x−b)2f'(x) = \frac{a-b}{(x-b)^2}