# Find the derivation of each of the following from the first principle:

Question:

Find the derivation of each of the following from the first principle:

$\sqrt{a x+b}$

Solution:

Let

$\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{ax}+\mathrm{b}}$

We need to find the derivative of $f(x)$ i.e. $f^{\prime}(x)$

We know that

$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{\mathrm{h}}$ …(i)

$f(x)=\sqrt{a x+b}$

$f(x+h)=\sqrt{a(x+h)+b}$

$=\sqrt{a x+a h+b}$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sqrt{a x+a h+b}-\sqrt{a x+b}}{h}$

Now rationalizing the numerator by multiplying and divide by the conjugate of

$\sqrt{a x+a h+b}-\sqrt{a x+b}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{a x+a h+b}-\sqrt{a x+b}}{h} \times \frac{\sqrt{a x+a h+b}+\sqrt{a x+b}}{\sqrt{a x+a h+b}+\sqrt{a x+b}}$

Using the formula:

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\lim _{h \rightarrow 0} \frac{(\sqrt{a x+a h+b})^{2}-(\sqrt{a x+b})^{2}}{h(\sqrt{a x+a h+b}+\sqrt{a x+b})}$

$=\lim _{h \rightarrow 0} \frac{a x+a h+b-a x-b}{h(\sqrt{a x+a h+b}+\sqrt{a x+b})}$

$=\lim _{h \rightarrow 0} \frac{a h}{h(\sqrt{a x+a h+b}+\sqrt{a x+b})}$

$=\lim _{h \rightarrow 0} \frac{a}{\sqrt{a x+a h+b}+\sqrt{a x+b}}$

Putting h = 0, we get

$=\frac{a}{\sqrt{a x+a(0)+b}+\sqrt{a x+b}}$

$=\frac{a}{\sqrt{a x+b}+\sqrt{a x+b}}$

$=\frac{a}{2 \sqrt{a x+b}}$

Hence,

$f^{\prime}(x)=\frac{a}{2 \sqrt{a x+b}}$