Question

Find the value of :
$\displaystyle \rm 1) \: \frac{d}{dx} x {}^{2}$
Also, Show how you got the result using the first principle method ​

Answer 1

Given :

f(x) = x²

To Find :

The dy/dx (or derivative) of f(x)

Solution :

According to the formula we know that :

$\displaystyle \rm f {}^{ \prime} ( {x}^{n} ) = n {x}^{n - 1}$

So, using the formula we get that :

$\implies\displaystyle \rm \frac{d}{dx} {x}^{2} = 2 {x}^{2 - 1}$

$\implies \: \displaystyle \rm \frac{d}{dx} {x}^{2} = 2x {}^{1}$

$\displaystyle \rm \implies \frac{d}{dx} {x}^{2} = 2x$

So, f'(x²) = 2x

Now, Using the First Principle Method:

We know that for differentiating any function w.r.t x , we use the first principle method :

$\to \boxed{\displaystyle \rm \lim_{h\to0} \frac{f(x + h) - f(x)}{h} }$

So, let's take a po.int x on the gra.ph of f(x) , then another po.int x+h with a change in value.

f(x) at x = x²

f(x) at x+h = (x+h)²

Now, using the F.P.M, we get that :

$\displaystyle \implies \rm\lim_{x\to 0} \frac{f(x + h) - f(x)}{h}$

$\implies \displaystyle \rm \lim_{x\to0} \frac{(x + h {)}^{2} - x {}^{2} }{h}$

${ \implies \displaystyle \rm\lim_{x\to0} \frac{(x + h + x) (x + h - x)}{h} }$

$\displaystyle \rm \implies \lim_{x\to0} \frac{(2x + h)(h)}{h}$

$\displaystyle \rm \implies \lim_{x\to0} (2x + h) \cancel{ \frac{h}{h}}$

$\displaystyle \rm \implies \lim_{x\to0}(2x + h)$

Now, Subt.stituting the value of h as 0, we get that :

$\displaystyle \rm \implies \frac{d}{dx} {x}^{2} = 2x$