questions

[bballj228]

Find the derivative. State the domain of the function and the domain of its derivative.

f(x) = x + √x

f(x) = (3 + x) / 1-3x

Find F'(a)

f(x) = (x^2 + 1) / (x - 2)

f(x) = √3x + 1

[Defennder]

And your work so far...?

[bballj228]

For the first one i got up to x + √x = x^1/2 = x^3/2

2nd one 1(1-3x) - (-3)(3 + x) all over (1-3x)^2 = 10 / (1 - 3x) ^2

3rd one i know its the quotient rule but not sure where to go with this one

For the fourth i tried the chain rule

√3x + 1 = 3x^1/2 + 1 = 1/2(3x)^1/2

[HallsofIvy]

For the first one i got up to x + √x = x^1/2 = x^3/2
? No, x+ \sqrt{x} is NOT equal to x^{1/2}= \sqrt{x} and NEITHER of those is equal to x^{1/2}. Did you mean that \sqrt{x}= x^{1/2}? And that the derivative is x^{3/2}? That last is not correct, either. Surely, you know what the derivative of x= x1 is 1? And what is the derivative of \sqrt{x}= x^{1/2}?

2nd one 1(1-3x) - (-3)(3 + x) all over (1-3x)^2 = 10 / (1 - 3x) ^2
Yes, that is correct!

3rd one i know its the quotient rule but not sure where to go with this one
USE the quotient rule of course! What is (x2+ 1)' ? What is (x- 2)'? Do it just like you did number 2.

For the fourth i tried the chain rule

√3x + 1 = 3x^1/2 + 1 = 1/2(3x)^1/2

Is that √(3)x+ 1, √(3x)+ 1, or √(3x+1)? In any case, none if those is equal to 3x^(1/2)+ 1.

The derivative of √(3) x+ 1 should be trivial. √(3x)+ 1 can be done as √(3)x^(1/2)+ 1, and √(3x+1) should be done using the chain rule: √(3x+1)= √u with u= 3x+1:
(√(3x+1))'= (du^(1/2)/du)(d(3x+1)/dx).

[bballj228]

[QUOTE=HallsofIvy;1777969]? No, x+ \sqrt{x} is NOT equal to x^{1/2}= \sqrt{x} and NEITHER of those is equal to x^{1/2}. Did you mean that \sqrt{x}= x^{1/2}? And that the derivative is x^{3/2}? That last is not correct, either. Surely, you know what the derivative of x= x1 is 1? And what is the derivative of \sqrt{x}= x^{1/2}?

No, I don't know actually.

[Defennder]

Use the power rule for differentiation. What is the derivative of x^n? Just apply it to x^1/2.