How do you use the limit definition of the derivative to find the derivative of #f(x)=sqrt(2x+7)#?

Question

Charles Anctil · Accepted Answer

# f'(x)=1/sqrt(2x+7)#.

Recall that, #f'(x)=lim_(t to x) (f(t)-f(x))/(t-x).# Since, #f(x)=sqrt(2x+7), :., f(t)=sqrt(2t+7).# # rArr f'(x)=lim_(t to x){sqrt(2t+7)-sqrt(2x+7)}/(t-x)# #=lim_(t to x){sqrt(2t+7)-sqrt(2x+7)}/(t-x)xx{sqrt(2t+7)+sqrt(2x+7)}/{sqrt(2t+7)+sqrt(2x+7)}# #=lim_(t to x){(2t+7)-(2x+7)}/{(t-x)(sqrt(2t+7)+sqrt(2x+7))# #=lim_(t to x){(2(t-x))/(t-x)}1/{sqrt(2t+7)+sqrt(2x+7)}# #=2/{sqrt(2x+7)+sqrt(2x+7)}# #:. f'(x)=1/sqrt(2x+7)#. Enjoy Maths.!

Christopher Agresta · Answer

undefined To use the limit definition of the derivative to find the derivative of $ f(x) = \sqrt{2x + 7} $, follow these steps:

1. Start with the limit definition of the derivative:
$$ f'(x) = \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h} $$

2. Substitute the given function $ f(x) = \sqrt{2x + 7} $ into the formula:
$$ f'(x) = \lim_{{h \to 0}} \frac{\sqrt{2(x + h) + 7} - \sqrt{2x + 7}}{h} $$

3. Rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator:
$$ f'(x) = \lim_{{h \to 0}} \frac{\sqrt{2(x + h) + 7} - \sqrt{2x + 7}}{h} \cdot \frac{\sqrt{2(x + h) + 7} + \sqrt{2x + 7}}{\sqrt{2(x + h) + 7} + \sqrt{2x + 7}} $$

4. Simplify the numerator and denominator:
$$ f'(x) = \lim_{{h \to 0}} \frac{(2(x + h) + 7) - (2x + 7)}{h(\sqrt{2(x + h) + 7} + \sqrt{2x + 7})} $$

5. Expand and simplify the numerator:
$$ f'(x) = \lim_{{h \to 0}} \frac{2x + 2h + 7 - 2x - 7}{h(\sqrt{2(x + h) + 7} + \sqrt{2x + 7})} $$
$$ f'(x) = \lim_{{h \to 0}} \frac{2h}{h(\sqrt{2(x + h) + 7} + \sqrt{2x + 7})} $$

6. Cancel out $ h $ in the numerator and denominator:
$$ f'(x) = \lim_{{h \to 0}} \frac{2}{\sqrt{2(x + h) + 7} + \sqrt{2x + 7}} $$

7. Evaluate the limit as $ h $ approaches 0:
$$ f'(x) = \frac{2}{\sqrt{2x + 7} + \sqrt{2x + 7}} $$
$$ f'(x) = \frac{2}{2\sqrt{2x + 7}} $$
$$ f'(x) = \frac{1}{\sqrt{2x + 7}} $$

So, the derivative of $ f(x) = \sqrt{2x + 7} $ is $ f'(x) = \frac{1}{\sqrt{2x + 7}} $.