Question

Given y=2x2-1, find dydx from first principles.

Answer 1

Given

y = 2x² - 1

We can write in this way too

f(x) = 2x² - 1

So,

f(x + h) = 2(x + h)² - 1

According to the first principle of derivative

$\sf \dfrac{dy}{dx} = \sf \lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{h}$

• substitute the value of f(x) and f(x+h)

$\sf \dfrac{dy}{dx} = \sf \lim_{h \rightarrow 0} \dfrac{[2(x + h)^{2} - 1]- [2x^{2} + 1 ]}{h}$

• Expand (x + h)² by using identity (a + b)²

$\sf \dfrac{dy}{dx} = \sf \lim_{h \rightarrow 0} \dfrac{[2(x^2 + h^2 + 2xh)- 1]- 2x^{2} + 1 }{h}$

• After expanding multiply from 2

$\sf \dfrac{dy}{dx} = \sf \lim_{h \rightarrow 0} \dfrac{[2x^2 + 2h^2 + 4xh - 1]- 2x^{2} + 1 }{h}$

• Simplify

$\sf \dfrac{dy}{dx} = \sf \lim_{h \rightarrow 0} \dfrac{ \cancel{2x^2}+ 2h^2 + 4xh - \cancel{1 } - \cancel{ 2x^{2}} + \cancel{1 }}{h}$

$\sf \dfrac{dy}{dx} = \sf \lim_{h \rightarrow 0} \dfrac{ 2h^2 + 4xh}{h}$

• Take h as a common and cancel it

$\sf \dfrac{dy}{dx} = \sf \lim_{h \rightarrow 0} \dfrac{ \cancel h(2h + 4x)}{ \cancel{h}}$

• Put the value of h

$\sf \dfrac{dy}{dx} = \sf \lim_{h \rightarrow 0} (2h + 4x)$

$\sf \dfrac{dy}{dx} =4x$

•°• The derivative of f(x) or y is 4x