Question 28.7.3: Differentiate: (i) (ax+b)^m(ii) (3x+5)⁶ (iii) √ax²+2bx+c...
Differentiate:
(i) (a x+b)^{m} \qquad(ii) (3 x+5)^{6} \qquad(iii) \sqrt{a x^{2}+2 b x+c}
(i) Let y=(a x+b)^{m} .
Put (a x+b)=t, \quad so that y=t^{m} and t=(a x+b) .
\therefore \frac{d y}{d t}=m t^{m-1} and \frac{d t}{d x}=a .
So, \frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right) \\=a m t^{m-1}=a m(a x+b)^{m-1} \quad[\because t=(a x+b)] .
(ii) Let y=(3 x+5)^{6} .
Put (3 x+5)=t , so that y=t^{6} and t=(3 x+5) .
\therefore \frac{d y}{d t}=6 t^{5} \text { and } \frac{d t}{d x}=3 \text {. }So, \frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right) \\=18 t^{5}=18(3 x+5)^{5} \quad[\because t=(3 x+5)] .
(iii) Let y=\sqrt{a x^{2}+2 b x+c} .
Put \left(a x^{2}+2 b x+c\right)=t , so that y=\sqrt{t} and t=\left(a x^{2}+2 b x+c\right) .
\therefore \frac{d y}{d t}=\frac{1}{2} t^{-1 / 2}=\frac{1}{2 \sqrt{t}} \text { and } \frac{d t}{d x}=(2 a x+2 b) \text {. }So, \frac{d y}{d x}=\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{1}{2 \sqrt{t}} \times 2(a x+b) \\ \\=\frac{(a x+b)}{\sqrt{t}}=\frac{(a x+b)}{\sqrt{a x^{2}+2 b x+c}} .