How do you use the limit definition to find the derivative of #f(x)=2/(5x+1)^3#?
By definition:
so:
regroup the binomials at the denominator
Perform the difference:
Expand now the power of the second binomial at the numerator:
The first two terms cancel each other:
The last two limits are zero, so:
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To find the derivative of ( f(x) = \frac{2}{(5x + 1)^3} ) using the limit definition:
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Start with the definition of the derivative: [ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]
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Substitute ( f(x) = \frac{2}{(5x + 1)^3} ) into the formula: [ f'(x) = \lim_{h \to 0} \frac{\frac{2}{(5(x+h) + 1)^3} - \frac{2}{(5x + 1)^3}}{h} ]
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Find a common denominator for the fractions: [ f'(x) = \lim_{h \to 0} \frac{2((5x + 1)^3 - (5(x+h) + 1)^3)}{h((5(x + h) + 1)^3 \cdot (5x + 1)^3)} ]
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Expand the binomials: [ f'(x) = \lim_{h \to 0} \frac{2(125x^3 + 75x^2 + 15x + 1 - (125(x^3 + 3x^2h + 3xh^2 + h^3) + 75(x^2 + 2xh + h^2) + 15(x + h) + 1))}{h((5(x + h) + 1)^3 \cdot (5x + 1)^3)} ]
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Simplify the expression: [ f'(x) = \lim_{h \to 0} \frac{2(125x^3 + 75x^2 + 15x + 1 - (125x^3 + 375x^2h + 375xh^2 + 125h^3 + 75x^2 + 150xh + 75h^2 + 15x + 15h + 1))}{h((5(x + h) + 1)^3 \cdot (5x + 1)^3)} ]
[ f'(x) = \lim_{h \to 0} \frac{2(75x^2 - 375x^2h + 15x - 150xh + 15h - 125h^2)}{h((5(x + h) + 1)^3 \cdot (5x + 1)^3)} ]
- Cancel out common terms: [ f'(x) = \lim_{h \to 0} \frac{2(-375x^2h - 150xh - 125h^2)}{h((5(x + h) + 1)^3 \cdot (5x + 1)^3)} ]
[ f'(x) = \lim_{h \to 0} \frac{-750x^2 - 300xh - 250h^2}{(5(x + h) + 1)^3 \cdot (5x + 1)^3} ]
- Apply the limit: [ f'(x) = \frac{-750x^2}{(5x + 1)^3 \cdot (5x + 1)^3} ]
[ f'(x) = \frac{-750x^2}{(5x + 1)^6} ]
Therefore, the derivative of ( f(x) = \frac{2}{(5x + 1)^3} ) is ( f'(x) = \frac{-750x^2}{(5x + 1)^6} ).
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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.
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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