Answer to Question #253739 in Calculus for Shambo

Question

Use the fact that lim
tà ƒ ƒ ¢ † ’0
e
t à ƒ ƒ ¢ ˆ ’ 1
t
= 1 to show from first principle that
d
dx(2e
2x à ƒ ƒ ¢ ˆ ’ e
à ƒ ƒ ¢ ˆ ’x + 5x) = 4e
2x + e
à ƒ ƒ ¢ ˆ ’x + 5.
[

Answer 1

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

$$=\lim\limits_{h\to0}\dfrac{2e^{2(x+h)}+e^{x+h}+5(x+h)-2e^{2x}-e^x-5x}{h}=$$

$$=\lim\limits_{h\to0}\dfrac{2e^{2x}(e^{2h}-1)}{h}+\lim\limits_{h\to0}\dfrac{e^{x}(e^{h}-1)}{h}+\lim\limits_{h\to0}\dfrac{5h}{h}$$

$$=4e^{2x}\lim\limits_{h\to0}\dfrac{e^{2h}-1}{2h}+e^x\lim\limits_{h\to0}\dfrac{e^h-1}{h}+5\lim\limits_{h\to0}\dfrac{h}{h}$$

$$=4e^{2x}(1)+e^x(1)+5(1)$$

$$=4e^{2x}+e^x+5$$