# The First Principles

[SOLVED] The First Principles

Hello!
This is my question:

1) Write down the definition of ƒ'(3), where $$f(x) = \sqrt{x + 1}$$

2) Use the definition to evaluate ƒ'(3).

This isn't an exercise from a book etc.. so I don't know what's the correct solution to this question but here's my attempt:

By the first principles( $$f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h}$$ ):
∴ $$\frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}$$

So would the definition of ƒ'(3) be like this:

$$f'(x) = \frac{\sqrt{3+h} - \sqrt{3+1}}{h}$$

Simplifying gives:
$$= 1$$

But another time I didn't use the first principles:

$$f(x) = \sqrt{x + 1}$$

$$1/2(x+1)^-1/2$$

$$\frac{1}{1/2.\sqrt{x+1}}$$

$$\frac{1}{1/2.\sqrt{3+1}} = 1$$

(?) - "?"

I got the same answer using both methods (=1). Am I on the right track? What else do I need to show?

Thank you. Last edited:

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Please recheck what you get for this:

$$f'(x) = \frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}$$

Remember you need to substitute x+h for x.

You mean like:
$$\frac{(x+h) - \sqrt{x+1}}{h}$$

$$\frac{(3+h) - \sqrt{3+1}}{h}$$

$$3 - 2$$ , → $$= 1$$

But is the rest of my above working correct? Did I show everything that was needed?
What would be the right solution to my question? Gib Z
Homework Helper
No thats not what he meant :( Recheck your working, if f(x) = sqrt( x+1), what is f(x+h) ? Its not sqrt(x+h).

HallsofIvy
Homework Helper
Hello!
This is my question:

1) Write down the definition of ƒ'(3), where $$f(x) = \sqrt{x + 1}$$

2) Use the definition to evaluate ƒ'(3).

This isn't an exercise from a book etc.. so I don't know what's the correct solution to this question but here's my attempt:

By the first principles( $$f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h}$$ ):
∴ $$\frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}$$

So would the definition of ƒ'(3) be like this:

$$f'(x) = \frac{\sqrt{3+h} - \sqrt{3+1}}{h}$$

No, for two reasons (1) you forgot the "+1" in the first term and (2) you forgot the limit !!
$$f'(x)= \lim_{h\rightarrow 0}\frac{\sqrt{3+h+1}-\sqrt{3+1}}{h}$$

Simplifying gives:
$$= 1$$
No, it does not! Even forgetting the "+1" $\sqrt{3+h}- \sqrt{3+1}$ is NOT equal to h and so the fraction is not equal to 1!

But another time I didn't use the first principles:

$$f(x) = \sqrt{x + 1}$$

$$1/2(x+1)^-1/2$$

$$\frac{1}{1/2.\sqrt{x+1}}$$
No,
$$\frac{1}{2\sqrt{x+1}}$$

$$\frac{1}{1/2.\sqrt{3+1}} = 1$$

(?) - "?"
$$\frac{1}{2\sqrt{3+1}}= \frac{1}{4}$$

I got the same answer using both methods (=1). Am I on the right track? What else do I need to show?

Thank you. 