# How do you find the derivative of #f (x) = 3x^5 + 4x# using the limit definition?

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To find the derivative of ( f(x) = 3x^5 + 4x ) using the limit definition:

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

Plug in ( f(x) = 3x^5 + 4x ):

[ f'(x) = \lim_{h \to 0} \frac{(3(x + h)^5 + 4(x + h)) - (3x^5 + 4x)}{h} ]

Expand and simplify:

[ f'(x) = \lim_{h \to 0} \frac{(3(x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5) + 4x + 4h) - (3x^5 + 4x)}{h} ]

[ f'(x) = \lim_{h \to 0} \frac{3x^5 + 15x^4h + 30x^3h^2 + 30x^2h^3 + 15xh^4 + 3h^5 + 4x + 4h - 3x^5 - 4x}{h} ]

[ f'(x) = \lim_{h \to 0} \frac{15x^4h + 30x^3h^2 + 30x^2h^3 + 15xh^4 + 3h^5 + 4h}{h} ]

[ f'(x) = \lim_{h \to 0} 15x^4 + 30x^3h + 30x^2h^2 + 15xh^3 + 3h^4 + 4 ]

[ f'(x) = 15x^4 + 4 ]

So, the derivative of ( f(x) = 3x^5 + 4x ) is ( f'(x) = 15x^4 + 4 ).

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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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