Definition of Derivative...4

[nycmathguy]

Calculus
Section 2.8

With question 32 we end Section 2.8 and Chapter 2.

 

[MathLover1]

(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,

 

[nycmathguy]

(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,
View attachment 3499

 

[MathLover1]

correct

 

[nycmathguy]

correct

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

 

[MathLover1]

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)

 

[nycmathguy]

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}

View attachment 3517


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

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

View attachment 3518

What a great reply. Thanks.

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

Definition of Derivative...2

 

[MathLover1]

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

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

 

[nycmathguy]

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?