Definition of Derivative...4

Calculus
Section 2.8





With question 32 we end Section 2.8 and Chapter 2.

 

(d/dx)(1/(1 + sqrt(x))). ......aply exponent rule 1/a=a^-1

=(d/dx)((1 + sqrt(x))^-1)..............apply chain rule

=-1/(1 + sqrt(x))^2*(d/dx)((1 + sqrt(x))............(d/dx)((1 + sqrt(x))=1/(2sqrt(x))

=-1/(1 + sqrt(x))^2*1/(2sqrt(x)).......simplify

=-1/((1 + sqrt(x))^2*2sqrt(x))

so,

 

correct

 

Can you find the range using a graph for g(x) and
g '(x)?

 

yes -> The range is the set of possible output values, which are shown on the y-axis.

f(x)=1/(1+sqrt(x))




range: { y element R : 0<y<=1}




f'(x)=-1/((1 + sqrt(x))^2*2sqrt(x))



range:{ y element R : y<0} (all negative real numbers)

 

What a great reply. Thanks.

How about finding the range for Definition of Derivatives...?

Definition of Derivative...2

 

finding the range for Definition of Derivatives????? what that means

you can find the derivative of the function using the definition of derivative

 

I meant to ask:

Can you please find the range for 26 and 28 located at the link provided?