Question

What is the instantaneous rate of change of `f(x)=sqrt(x^2+4x+2)` at `x=0`?

Answer 1

`sqrt2`

Explanation:

The instantaneous rate of change is the value of f'(0).

`f(x)=sqrt(x^2+4x+2)=(x^2+4x+2)^(1/2)`

differentiate using the`color(blue)" chain rule"`

`d/dx[f(g(x))]=f'(g(x)).g'(x)"......(A)"`
`"-------------------------------------------------------"`

`f(g(x))=(x^2+4x+2)^(1/2)`

`rArrf'(g(x))=1/2(x^2+4x+2)^(-1/2)`

and `g(x)=x^2+4x+2rArrg'(x)=2x+4`
`"-----------------------------------------------------------"`
Substitute these values into (A)

`rArrf'(x)=1/2(x^2+4x+2)^(-1/2) .(2x+4)`

`=1/2xx1/(x^2+4x+2)^(1/2)xx(2x+4)`

`rArrf'(0)=1/2xx1/sqrt2xx4=2/sqrt2=(2sqrt2)/2=sqrt2`