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

1 Answer
Jim G.
May 17, 2016

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 thecolor(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

Related questions
See all questions in Instantaneous Rate of Change at a Point
Impact of this question
1793 views around the world
You can reuse this answer
Creative Commons License