Question

find the derivatives using defination​

Answer 1

Answer:

$\huge\boxed{(i)\ f'(x)=10x}\\\boxed{(ii)\ f'(x)=\dfrac{1}{2\sqrt{x+1}}}$

Step-by-step explanation:

$f'(x)=\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}$

$(i)\ f(x)=5x^2-3\\\\f'(x)=\lim\limits_{h\to0}\dfrac{5(x+h)^2-3-(5x^2-3)}{h}\\\\=\lim\limits_{h\to0}\dfrac{5(x^2+2hx+h^2)-3-5x^2+3}{h}\\\\=\lim\limits_{h\to0}\dfrac{5x^2+10hx+5h^2-5x^2}{h}\\\\=\lim\limits_{h\to0}\dfrac{10hx+5h^2}{h}\\\\=\lim\limits_{h\to0}\dfrac{h\!\!\!\!\diagup(10x+5h)}{h\!\!\!\!\diagup}\\\\=\lim\limits_{h\to0}(10x+5h)=10x+5\cdot0=10x$

$(ii)\ f(x)=\sqrt{x+1}\\\\f'(x)=\lim\limits_{h\to0}\dfrac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\\\\=\lim\limits_{h\to0}\dfrac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\cdot\dfrac{\sqrt{x+h+1}+\sqrt{x+1}}{\sqrt{x+h+1}+\sqrt{x+1}}\\\\=\lim\limits_{h\to0}\dfrac{(\sqrt{x+h+1})^2-(\sqrt{x+1})^2}{h(\sqrt{x+h+1}+\sqrt{x+1})}\\\\=\lim\limits_{h\to0}\dfrac{x+h+1-(x+1)}{h(\sqrt{x+h+1}+\sqrt{x+1})}\\\\=\lim\limits_{h\to0}\dfrac{x+h+1-x-1}{h(\sqrt{x+h+1}+\sqrt{x+1})}$

$=\lim\limits_{h\to0}\dfrac{h\!\!\!\!\diagup}{h\!\!\!\!\diagup(\sqrt{x+h+1}+\sqrt{x+1})}\\\\=\lim\limits_{h\to0}\dfrac{1}{\sqrt{x+h+1}+\sqrt{x+1}}\\\\=\dfrac{1}{\sqrt{x+0+1}+\sqrt{x+1}}\\\\=\dfrac{1}{\sqrt{x+1}+\sqrt{x+1}}\\\\=\dfrac{1}{2\sqrt{x+1}}$