How do you use the limit definition of the derivative to find the derivative of #f(x)=x^3+2x#?

Answer 1

#d/dx (x^3+2x) = 3x^2+2#

By definition:

#(df)/dx = lim_(h->0) (f(x+h)-f(x))/h#

so:

#d/dx (x^3+2x) = lim_(h->0) ((x+h)^3+2(x+h)-x^3-2x)/h#

Expand the cube of the binomial and simplify:

#d/dx (x^3+2x) = lim_(h->0) (color(blue)(x^3)+3x^2h+3xh^2+h^3 +color(red)(2x)+2h-color(blue)(x^3)-color(red)(2x))/h#
#d/dx (x^3+2x) = lim_(h->0) (3x^2h+3xh^2+h^3 +2h)/h#
#d/dx (x^3+2x) = lim_(h->0) (cancel(h)(3x^2+3xh+h^2 +2))/cancel(h)#
#d/dx (x^3+2x) = lim_(h->0) 3x^2+3xh+h^2 +2#
#d/dx (x^3+2x) = 3x^2+2#
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Answer 2

To find the derivative of ( f(x) = x^3 + 2x ) using the limit definition of the derivative, follow these steps:

  1. Write down the limit definition of the derivative: [ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

  2. Substitute ( f(x) = x^3 + 2x ) into the formula: [ f'(x) = \lim_{h \to 0} \frac{(x + h)^3 + 2(x + h) - (x^3 + 2x)}{h} ]

  3. Expand ( (x + h)^3 ) and simplify: [ (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 ] [ f'(x) = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 + 2x + 2h - x^3 - 2x}{h} ]

  4. Cancel out like terms: [ f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 + 2h}{h} ]

  5. Factor out an ( h ) from the numerator: [ f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2) + 2h}{h} ]

  6. Cancel out an ( h ) from the numerator and denominator: [ f'(x) = \lim_{h \to 0} 3x^2 + 3xh + h^2 + 2 ]

  7. Evaluate the limit as ( h ) approaches 0: [ f'(x) = 3x^2 + 2 ]

So, the derivative of ( f(x) = x^3 + 2x ) is ( f'(x) = 3x^2 + 2 ).

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Answer 3

To use the limit definition of the derivative to find the derivative of ( f(x) = x^3 + 2x ), follow these steps:

  1. Write down the limit definition of the derivative:

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

  1. Substitute the function ( f(x) = x^3 + 2x ) into the definition:

[ f'(x) = \lim_{h \to 0} \frac{(x+h)^3 + 2(x+h) - (x^3 + 2x)}{h} ]

  1. Expand ( (x+h)^3 ) using the binomial theorem:

[ (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 ]

  1. Substitute the expanded expression into the limit expression:

[ f'(x) = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 + 2(x+h) - (x^3 + 2x)}{h} ]

  1. Simplify the expression by canceling out common terms:

[ f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 + 2h}{h} ]

  1. Factor out ( h ) from the numerator:

[ f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2) + 2h}{h} ]

  1. Cancel out ( h ) from the numerator and denominator:

[ f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2 + 2) ]

  1. Now, substitute ( h = 0 ) into the expression:

[ f'(x) = 3x^2 + 2 ]

So, the derivative of ( f(x) = x^3 + 2x ) is ( f'(x) = 3x^2 + 2 ).

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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.

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.

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.

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.

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