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**[SOLVED] The First Principles**

**Hello!**

This is my question:

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

This is my question:

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

**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( [tex]f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h}[/tex] ):

∴ [tex]\frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}[/tex]

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

[tex]f'(x) = \frac{\sqrt{3+h} - \sqrt{3+1}}{h}[/tex]

Simplifying gives:

[tex]= 1[/tex]

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

[tex]f(x) = \sqrt{x + 1}[/tex]

[tex]1/2(x+1)^-1/2[/tex]

[tex]\frac{1}{1/2.\sqrt{x+1}}[/tex]

[tex]\frac{1}{1/2.\sqrt{3+1}} = 1[/tex]

(?) - "?"

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

Thank you.

By the first principles( [tex]f'(x) = lim h\rightarrow0 \frac{f(x+h) - f(x)}{h}[/tex] ):

∴ [tex]\frac{\sqrt{(x+h)} - \sqrt{x+1}}{h}[/tex]

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

[tex]f'(x) = \frac{\sqrt{3+h} - \sqrt{3+1}}{h}[/tex]

Simplifying gives:

[tex]= 1[/tex]

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

[tex]f(x) = \sqrt{x + 1}[/tex]

[tex]1/2(x+1)^-1/2[/tex]

[tex]\frac{1}{1/2.\sqrt{x+1}}[/tex]

[tex]\frac{1}{1/2.\sqrt{3+1}} = 1[/tex]

(?) - "?"

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

Thank you.

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