Question

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

Answer 1

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

Explanation:

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.!