It follows from first principles that the derivative of a function f ( x ) f(x) f ( x )
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x )
In our case, for f ( x ) = 1 f(x)=1 f ( x ) = 1
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 ( 1 ) − 1 h = = lim h → 0 ( 0 h ) = 0 → f ( x ) = 1 → f ′ ( x ) = 0 f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\frac{(1)-1}{h}=\\[0.3cm]
=\lim\limits_{h\to0}\left(\frac{0}{h}\right)=0\to\boxed{f(x)=1\to f'(x)=0} f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) = h → 0 lim h ( 1 ) − 1 = = h → 0 lim ( h 0 ) = 0 → f ( x ) = 1 → f ′ ( x ) = 0
For f ( x ) = x + 1 f(x)=\sqrt{x+1} f ( x ) = x + 1
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 x + 1 + h − x + 1 h = = lim h → 0 ( x + 1 + h − x + 1 ) ( x + 1 + h + x + 1 ) h ( x + 1 + h + x + 1 ) = = lim h → 0 ( x + 1 + h ) − ( x + 1 ) h ( x + 1 + h + x + 1 ) = = lim h → 0 h h ( x + 1 + h + x + 1 ) = = lim h → 0 1 ( x + 1 + h + x + 1 ) = 1 2 x + 1 f ( x ) = x + 1 → f ′ ( x ) = 1 2 x + 1 f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\frac{\sqrt{x+1+h}-\sqrt{x+1}}{h}=\\[0.3cm]
=\lim\limits_{h\to0}\frac{\left(\sqrt{x+1+h}-\sqrt{x+1}\right)\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}{h\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}=\\[0.3cm]
=\lim\limits_{h\to0}\frac{\left(x+1+h\right)-\left(x+1\right)}{h\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}=\\[0.3cm]
=\lim\limits_{h\to0}\frac{h}{h\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}=\\[0.3cm]
=\lim\limits_{h\to0}\frac{1}{\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}=\frac{1}{2\sqrt{x+1}}\\[0.3cm]
\boxed{f(x)=\sqrt{x+1}\to f'(x)=\frac{1}{2\sqrt{x+1}}} f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) = h → 0 lim h x + 1 + h − x + 1 = = h → 0 lim h ( x + 1 + h + x + 1 ) ( x + 1 + h − x + 1 ) ( x + 1 + h + x + 1 ) = = h → 0 lim h ( x + 1 + h + x + 1 ) ( x + 1 + h ) − ( x + 1 ) = = h → 0 lim h ( x + 1 + h + x + 1 ) h = = h → 0 lim ( x + 1 + h + x + 1 ) 1 = 2 x + 1 1 f ( x ) = x + 1 → f ′ ( x ) = 2 x + 1 1
#340153 on Dec 2023