Question

Do anyone know what is the answer

Answer 1

"first principle" means "use the limit definition of the derivative":

`f'(x) = \displaystyle \lim_(h\to0) \frac{f(x+h) - f(x)}h`

For f(x) = y = ax² + b, the derivative is

`f'(x) = \displaystyle \lim_(h\to0) \frac{(a(x+h)^2+b) - (ax^2+b)}h`

`f'(x) = \displaystyle \lim_(h\to0) \frac{a(x^2+2xh+h^2) - ax^2}h`

`f'(x) = \displaystyle \lim_(h\to0) \frac{2axh+ah^2}h`

`f'(x) = \displaystyle \lim_(h\to0) (2ax+ah) = \boxed{2ax}`

Answer 2

We are given with a function y = ax² + b and are asked to find it's derivative by first principle of differentiation , so by first principle we know that :-

• 
  `{\boxed{\displaystyle \bf f^(\prime)(x)=\lim_(h\to 0)(f(x+h)-f(x))/(h)}}`

Now , here f(x) is y = ax² + b , so , f(x+h) will be a(x+h)²+b . Now by first principle ;

`{:\implies \quad \displaystyle \sf y^(\prime)=\lim_(h\to 0)\frac{\{a(x+h)^(2)+b\}-(ax^(2)+b)}{h}}`

`{:\implies \quad \displaystyle \sf y^(\prime)=\lim_(h\to 0)\frac{\{a(x^(2)+h^(2)+2xh)+b\}-(ax^(2)+b)}{h}\quad \qquad \{\because (a+b)^(2)=a^(2)+b^(2)+2ab\}}`

`{:\implies \quad \displaystyle \sf y^(\prime)=\lim_(h\to 0)\frac{\cancel{ax^(2)}+ah^(2)+2axh+\cancel{b}-\cancel{ax^(2)}-\cancel{b}}{h}}`

`{:\implies \quad \displaystyle \sf y^(\prime)=\lim_(h\to 0)\frac{\cancel{h}(ah+2ax)}{\cancel{h}}}`

`{:\implies \quad \displaystyle \sf y^(\prime)=a* 0+2ax}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{y^(\prime)=2ax}}}`