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

Solution:

Let $f(x)=x^{3}-2 x^{2}+x+3$

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)

$f(x)=x^{3}-2 x^{2}+x+3$

$f(x+h)=(x+h)^{3}-2(x+h)^{2}+(x+h)+3$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{3}-2(x+h)^{2}+(x+h)+3-\left[x^{3}-2 x^{2}+x+3\right]}{h}$

$=\lim _{h \rightarrow 0} \frac{(x+h)^{3}-2(x+h)^{2}+(x+h)+3-x^{3}+2 x^{2}-x-3}{h}$

$=\lim _{h \rightarrow 0} \frac{\left[(x+h)^{3}-x^{3}\right]-2\left[(x+h)^{2}-x^{2}\right]+[x+h-x]}{h}$

Using the identities:

$(a+b)^{3}=a^{3}+b^{3}+3 a b^{2}+3 a^{2} b$

$(a+b)^{2}=a^{2}+b^{2}+2 a b$

$=\lim _{h \rightarrow 0} \frac{\left[x^{3}+h^{3}+3 x h^{2}+3 x^{2} h-x^{3}\right]-2\left[x^{2}+h^{2}+2 x h-x^{2}\right]+h}{h}$

$=\lim _{h \rightarrow 0} \frac{\left[h^{3}+3 x h^{2}+3 x^{2} h\right]-2\left[h^{2}+2 x h\right]+h}{h}$

$=\lim _{h \rightarrow 0} \frac{h\left[h^{2}+3 x h+3 x^{2}\right]-2 h[h+2 x]+h}{h}$

$=\lim _{h \rightarrow 0} h^{2}+3 x h+3 x^{2}-2 h-4 x+1$

Putting h = 0, we get

$f^{\prime}(x)=(0)^{2}+2 x(0)+3 x^{2}-2(0)-4 x+1$

$=3 x^{2}-4 x+1$

Hence, $f^{\prime}(x)=3 x^{2}-4 x+1$