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

Answer 1

Evelyn Agard

#f'(x)=2x+1#

#lim_(hrarr0)[[(x+h)^2+x+h]-[x^2+x]]/h#

#lim_(hrarr0)[[x^2+2xh+h^2+x+h]-[x^2+x]]/h#

#lim_(hrarr0)[x^2+2xh+h^2+x+h-x^2-x]/h#

#lim_(hrarr0)[2xh+h^2+h]/h#

#lim_(hrarr0)[h(2x+h+1)]/h#

#lim_(hrarr0)2x+h+1=2x+0+1=2x+1#

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Answer 2

Adam Adkins

To find ( f'(x) ) using the definition of a derivative for ( f(x) = x^2 + x ), we first recall that the definition of the derivative of a function ( f(x) ) at a point ( x ) is given by:

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

Substituting ( f(x) = x^2 + x ) into this formula, we have:

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

Expanding ( (x + h)^2 ) and simplifying the expression, we get:

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

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

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

Now, as ( h ) approaches 0, the term ( 2x ) remains constant, and ( h + 1 ) approaches 1. Therefore, the derivative ( f'(x) ) is:

[ f'(x) = 2x + 1 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.