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

Solution:

 Let

$f(x)=\sqrt{5 x-4}$

We need to find the derivative of f(x) i.e. f’(x)

We know that,

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

$f(x)=\sqrt{5 x-4}$

$f(x+h)=\sqrt{5(x+h)-4}$

$=\sqrt{5 x+5 h-4}$

Putting values in (i), we get

$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\sqrt{5 \mathrm{x}+5 \mathrm{~h}-4}-\sqrt{5 \mathrm{x}-4}}{\mathrm{~h}}$

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

$\sqrt{5 x+5 h-4}-\sqrt{5 x-4}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{5 x+5 h-4}-\sqrt{5 x-4}}{h} \times \frac{\sqrt{5 x+5 h-4}+\sqrt{5 x-4}}{\sqrt{5 x+5 h-4}+\sqrt{5 x-4}}$

Using the formula:

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

$=\lim _{h \rightarrow 0} \frac{(\sqrt{5 x+5 h-4})^{2}-(\sqrt{5 x-4})^{2}}{h(\sqrt{5 x+5 h-4}+\sqrt{5 x-4})}$

$=\lim _{h \rightarrow 0} \frac{5 x+5 h-4-5 x+4}{h(\sqrt{5 x+5 h-4}+\sqrt{5 x-4})}$

$=\lim _{h \rightarrow 0} \frac{5 h}{h(\sqrt{5 x+5 h-4}+\sqrt{5 x-4})}$

$=\lim _{h \rightarrow 0} \frac{5}{\sqrt{5 x+5 h-4}+\sqrt{5 x-4}}$

Putting h = 0, we get

$=\frac{5}{\sqrt{5 x+5(0)-4}+\sqrt{5 x-4}}$

$=\frac{5}{\sqrt{5 x-4}+\sqrt{5 x-4}}$

$=\frac{5}{2 \sqrt{5 x-4}}$

Hence,

$f^{\prime}(x)=\frac{5}{2 \sqrt{5 x-4}}$