# How do you find f'(x) using the definition of a derivative #f(x) =(x^2 + 2)^2#?

There are two forms of the limit definition of the derivative.

For this problem, the first form is easier to use.

Which is the same result obtained from later techniques such as the power rule and chain rule.

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To find ( f'(x) ) using the definition of a derivative for the function ( f(x) = (x^2 + 2)^2 ), we start by applying the definition:

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

Substitute ( f(x) = (x^2 + 2)^2 ) into the formula:

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

Expand ( (x+h)^2 ) and ( (x^2 + 2)^2 ):

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

Now, subtract the squared terms:

[ f'(x) = \lim_{h \to 0} \frac{(x^2 + 2h + h^2 + 2)^2 - (x^2 + 2)^2}{h} ] [ = \lim_{h \to 0} \frac{(x^2 + 2h + h^2 + 2 + x^2 + 2)^2 - (x^2 + 2)^2}{h} ] [ = \lim_{h \to 0} \frac{(x^2 + 2h + h^2 + 2x^2 + 4h + 4 + 4x^2 + 4h + 4 + h^2)^2 - (x^2 + 2)^2}{h} ]

Simplify and expand the terms within the limit:

[ f'(x) = \lim_{h \to 0} \frac{6x^2h + 6h^2 + 8h}{h} ]

Now, cancel out the ( h ) in the numerator and denominator:

[ f'(x) = \lim_{h \to 0} (6x^2 + 6h + 8) ]

Take the limit as ( h ) approaches 0:

[ f'(x) = 6x^2 + 8 ]

So, the derivative of ( f(x) = (x^2 + 2)^2 ) with respect to ( x ) is ( f'(x) = 6x^2 + 8 ).

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