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This a demonstration of two ways of finding the derivative of a function {$f(x)$} at a point {$(x_0,f(x_0))$}.

For this demonstration, the function that will be used is

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

and the point used will be {$(3,2)$}, which can be verified to be a point on the function.

We use the basic definition of the derivative:

{$$ f^\prime (x) = \lim_{ h \rightarrow0} {{f(x + h) -f(x)} \over { h}} $$}

But at the particular point, which we know to be on the function, we know {$ x=3 $} and {$ f(x) = 2). As in most case, it is easier to express a root as an exponent, so

{$$ f(3+h) = \sqrt{(3+h)+1} = ((3+h) + 1)^{1 \over 2} = (4+h)^{1\over 2} $$}

{$$ \begin{align} f^\prime (3) &= \lim_{ h \rightarrow0} {{f(3 + h) -f(3)} \over { h}} \cr &= \lim_{ h \rightarrow0} {{(4+h)^{1\over 2} -2} \over { h}} \cr &= \lim_{ h \rightarrow0} {{[(4+h)^{1\over 2} -2][(4+h)^{1\over 2} +2]} \over { h[(4+h)^{1\over 2} +2]}} \cr &= \lim_{ h \rightarrow0} {{[(4+h)^{1\over 2}]^2 -4} \over { h[(4+h)^{1\over 2} +2]}} \cr &= \lim_{ h \rightarrow0} {{(4+h) -4} \over { h[(4+h)^{1\over 2} +2]}} \cr &= \lim_{ h \rightarrow0} {h \over { h[(4+h)^{1\over 2} +2]}} \cr &= \lim_{ h \rightarrow0} {1 \over { (4+h)^{1\over 2} +2}} \cr &= {1 \over { (4+0)^{1\over 2} +2}} \cr f^\prime (3) &= {1 \over 4} \end{align} $$}

The first method may be a large amount of work to invest on just one point of just one function. You may know a general formula for finding the derivative function quickly.

Powers Rule

{$$ {d \over dx} x^n = nx^{n-1} $$}

With the power rule it is very easy to find the function for the derivative.

{$$ f(x) = \sqrt{x+1} = (x+1)^{1 \over 2} $$}

so letting {$ n = {1 \over 2}$} and applying the powers rule:

{$$ f'(x) = {1 \over 2} (x+1)^{{1 \over 2} -1} = {1 \over 2} (x+1)^{-{1\over 2}} $$}

Then by substituting the value of *x* at (3,2) in the derivative function, the derivative at the point is found:

{$$ f'(3) = {1 \over 2} (3+1)^{-{1\over 2}} = ({1 \over 2})({1 \over 2}) = {1 \over 4} $$}

However, the derivative function can also be obtained directly from the definition of derivative.

{$$ \begin{align} f^\prime (x) &= \lim_{ h \rightarrow0} {{f(x + h) -f(x)} \over { h}} \cr &= \lim_{ h \rightarrow0} {{[(x + h) + 1]^{1 \over 2} - [x+1]^{1 \over 2}} \over { h}} \cr &= \lim_{ h \rightarrow0} {{\{[(x + h) + 1]^{1 \over 2} - [x+1]^{1 \over 2}\}\{[(x + h) + 1]^{1 \over 2} + [x+1]^{1 \over 2}\}} \over { (h)\{[(x + h) + 1]^{1 \over 2} + [x+1]^{1 \over 2}\}}} \cr &= \lim_{ h \rightarrow0} {{\{[(x + h) + 1]^{1 \over 2}\}^2 - \{[x+1]^{1 \over 2}\}^2} \over { (h)\{[(x + h) + 1]^{1 \over 2} + [x+1]^{1 \over 2}\}}} \cr &= \lim_{ h \rightarrow0} {{\{x + h + 1\} - \{x+1\}} \over { (h)\{[(x + h) + 1]^{1 \over 2} + [x+1]^{1 \over 2}\}}} \cr &= \lim_{ h \rightarrow0} {h \over { (h)\{[(x + h) + 1]^{1 \over 2} + [x+1]^{1 \over 2}\}}} \cr &= \lim_{ h \rightarrow0} {1 \over { [(x + h) + 1]^{1 \over 2} + [x+1]^{1 \over 2}}} \cr &= {1 \over { (x + 1)^{1 \over 2} + (x+1)^{1 \over 2}}} \cr f^\prime (x) &= {1 \over { 2(x + 1)^{1 \over 2}}} = {1 \over {2 \sqrt{x+1} } } \end{align} $$}

Then the *x* value of the point can be plugged in to find the derivative at the point.

{$$ f^\prime (3) = {1 \over {2 \sqrt{3+1} } }= {1 \over 4} $$}

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