# How do you find the derivative of #f(x)=sqrt(x+1)# using the limit process?

Find a substitution for h that cancels the numerator and allows the limit to go to zero.

For integer roots of any function, we can find a substitution for h:

Square both sides:

Regroup the right side:

This factors:

When we make the substitution for h, the numerator will cancel the left factor:

Here is the limit with the canceled factors eliminated:

We can let the limit to go zero:

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To find the derivative of ( f(x) = \sqrt{x+1} ) using the limit process, we apply the definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} ]

First, substitute ( f(x) = \sqrt{x+1} ) into the definition:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+1} - \sqrt{x+1}}{h} ]

To eliminate the square roots, we use the conjugate technique by multiplying both the numerator and denominator by the conjugate of the numerator:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+1} - \sqrt{x+1}}{h} \times \frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}} ]

This simplifies to:

[ f'(x) = \lim_{h \to 0} \frac{(x+h+1) - (x+1)}{h(\sqrt{x+h+1} + \sqrt{x+1})} ]

[ f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h+1} + \sqrt{x+1})} ]

[ f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}} ]

Now, as ( h ) approaches 0, the expression becomes:

[ f'(x) = \frac{1}{\sqrt{x+1} + \sqrt{x+1}} ]

[ f'(x) = \frac{1}{2\sqrt{x+1}} ]

Therefore, the derivative of ( f(x) = \sqrt{x+1} ) using the limit process is ( f'(x) = \frac{1}{2\sqrt{x+1}} ).

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To find the derivative of ( f(x) = \sqrt{x + 1} ) using the limit process, you first express the function as ( f(x) = (x + 1)^{\frac{1}{2}} ). Then, apply the definition of the derivative:

[ f'(x) = \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h} ]

Substitute ( f(x) = (x + 1)^{\frac{1}{2}} ) into the formula:

[ f'(x) = \lim_{{h \to 0}} \frac{(x + h + 1)^{\frac{1}{2}} - (x + 1)^{\frac{1}{2}}}{h} ]

Now, to simplify the expression inside the limit, we use the conjugate method:

[ \frac{(x + h + 1)^{\frac{1}{2}} - (x + 1)^{\frac{1}{2}}}{h} \cdot \frac{(x + h + 1)^{\frac{1}{2}} + (x + 1)^{\frac{1}{2}}}{(x + h + 1)^{\frac{1}{2}} + (x + 1)^{\frac{1}{2}}} ]

This gives us:

[ f'(x) = \lim_{{h \to 0}} \frac{(x + h + 1) - (x + 1)}{h((x + h + 1)^{\frac{1}{2}} + (x + 1)^{\frac{1}{2}})} ]

Simplifying further:

[ f'(x) = \lim_{{h \to 0}} \frac{h}{h((x + h + 1)^{\frac{1}{2}} + (x + 1)^{\frac{1}{2}})} ]

Cancel out the common factor of ( h ):

[ f'(x) = \lim_{{h \to 0}} \frac{1}{(x + h + 1)^{\frac{1}{2}} + (x + 1)^{\frac{1}{2}}} ]

Now, substitute ( h = 0 ) into the limit:

[ f'(x) = \frac{1}{(x + 1)^{\frac{1}{2}} + (x + 1)^{\frac{1}{2}}} ]

Finally, simplify the denominator:

[ f'(x) = \frac{1}{2(x + 1)^{\frac{1}{2}}} ]

Thus, the derivative of ( f(x) = \sqrt{x + 1} ) using the limit process is ( f'(x) = \frac{1}{2(x + 1)^{\frac{1}{2}}} ).

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