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 :-

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

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

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

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

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

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

${:\implies \quad \displaystyle \sf y^{\prime}=a\times 0+2ax}$

${:\implies \quad \bf \therefore \quad \underline{\underline{y^{\prime}=2ax}}}$