Question

For the function `f(x) = sqrt(x^2+49)`, what is `f''(x)`. What is `f''(0)` and `f''(9)`?

Answer 1

`f''(0)=1/7` and `f''(9)=0.0816`

Explanation:

`f''(x)` is the second derivative and `f'(x)` is first derivative.

As `f(x)=sqrt(x^2+49)`

`f'(x)=1/(2sqrt(x^2+49))xx2x=x/sqrt(x^2+49)`

and `f''(x)=(sqrt(x^2+49)-x/(sqrt(x^2+49)))/(x^2+49)`

= `(x^2+49-x)/(x^2+49)^(3/2)=(x^2-x+49)/(x^2+49)^(3/2)`

Hence, `f''(0)=49/49^(3/2)=49/343=1/7`

and `f''(9)=(81-9+49)/(130)^(3/2)=121/130^(3/2)=0.0816`