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secondary 4 | A Maths
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I rarely see this kind of ln, please teach me on how to differenciate this kind of ln. Thx in advance.
Date Posted: 3 years ago
Views: 230
I think I mentioned before that your lecturer was using an older/alternative style of notation for ln (which is not used at Singapore-Cambridge O or A levels)
Anyway,
ln³ 5x just means (ln 5x)³, so we differentiate like how we do usually, using the chain rule.
i.e bring the power down, subtract one from the power and then multiply by the derivative of the inside term.
d/dx [f(x)]ⁿ = n · [f(x)]ⁿ-¹ · f'(x)
y = ln³ 5x = (ln 5x)³
dy/dx = 3 (ln 5x)² (5/5x)
= 3 (ln 5x)² / x
(Or 3ln²5x / x , as used in your lecturer's notation)
ln³ 5x just means (ln 5x)³, so we differentiate like how we do usually, using the chain rule.
i.e bring the power down, subtract one from the power and then multiply by the derivative of the inside term.
d/dx [f(x)]ⁿ = n · [f(x)]ⁿ-¹ · f'(x)
y = ln³ 5x = (ln 5x)³
dy/dx = 3 (ln 5x)² (5/5x)
= 3 (ln 5x)² / x
(Or 3ln²5x / x , as used in your lecturer's notation)
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Check your question's comments section please...
Thanks.
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