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NCERT Exemplar · Q44

Q.Differentiate with respect to xx using first principle: ax+bcx+d\dfrac{ax + b}{cx + d}.

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The derivative of ax+bcx+d\frac{ax+b}{cx+d} is found using the first principle of differentiation (limit definition). The final result is ad−bc(cx+d)2\frac{ad - bc}{(cx+d)^2}.

The first principle — also called the definition of the derivative — asks us to compute the limit of the difference quotient as the change in xx approaches zero. For a function f(x)f(x), this is:

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

Here, f(x)=ax+bcx+df(x) = \frac{ax+b}{cx+d}. The key challenge is algebraic: we need to combine the two fractions in the numerator, simplify, and then take the limit. The quotient rule is a shortcut for this, but using first principles directly builds a deeper understanding of why the derivative formula works.

Let’s go step by step.

  1. Write the difference quotient. We need:

f(x+h)−f(x)h=a(x+h)+bc(x+h)+d−ax+bcx+dh.\frac{f(x+h) - f(x)}{h} = \frac{ \frac{a(x+h)+b}{c(x+h)+d} - \frac{ax+b}{cx+d} }{h}.

  1. Combine the two fractions in the numerator. The common denominator is (c(x+h)+d)(cx+d)(c(x+h)+d)(cx+d). So:

a(x+h)+bc(x+h)+d−ax+bcx+d=[a(x+h)+b](cx+d)−(ax+b)[c(x+h)+d](c(x+h)+d)(cx+d).\frac{a(x+h)+b}{c(x+h)+d} - \frac{ax+b}{cx+d} = \frac{ [a(x+h)+b](cx+d) - (ax+b)[c(x+h)+d] }{ (c(x+h)+d)(cx+d) }.

  1. Expand the numerator carefully.

    First term: [a(x+h)+b](cx+d)=a(x+h)(cx+d)+b(cx+d)[a(x+h)+b](cx+d) = a(x+h)(cx+d) + b(cx+d).

    Second term: (ax+b)[c(x+h)+d]=(ax+b)[c(x+h)]+(ax+b)d(ax+b)[c(x+h)+d] = (ax+b)[c(x+h)] + (ax+b)d.

    It’s easier to expand systematically:

    • [a(x+h)+b](cx+d)=a(x+h)(cx)+a(x+h)d+b(cx)+bd[a(x+h)+b](cx+d) = a(x+h)(cx) + a(x+h)d + b(cx) + bd =acx(x+h)+ad(x+h)+bcx+bd= a c x(x+h) + a d (x+h) + b c x + b d.
    • (ax+b)[c(x+h)+d]=ax⋅c(x+h)+ax⋅d+b⋅c(x+h)+b⋅d(ax+b)[c(x+h)+d] = ax \cdot c(x+h) + ax \cdot d + b \cdot c(x+h) + b \cdot d =acx(x+h)+adx+bc(x+h)+bd= a c x (x+h) + a d x + b c (x+h) + b d.

    Now subtract the second from the first:

Numerator=[acx(x+h)+ad(x+h)+bcx+bd]−[acx(x+h)+adx+bc(x+h)+bd].\text{Numerator} = [a c x(x+h) + a d (x+h) + b c x + b d] - [a c x(x+h) + a d x + b c (x+h) + b d].

Notice acx(x+h)a c x(x+h) cancels, and bdb d cancels. We are left with:

ad(x+h)−adx+bcx−bc(x+h)=adh+bc(x−(x+h))=adh−bch=h(ad−bc).a d (x+h) - a d x + b c x - b c (x+h) = a d h + b c (x - (x+h)) = a d h - b c h = h(ad - bc). …

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