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Siazzle
@Siazzle
May 2019
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Bonjour! Pouvez-vous m’aider à dériver cette fonction svp ?
f(x) = (1+lnx)^2 -2x
f’(x) = ?
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scoladan
Verified answer
Bonjour,
(1 + lnx)² de la forme uⁿ donc dérivée : nu'uⁿ⁻¹ soit 2(1 + lnx)/x
⇒ f'(x) = 2(1 + lnx)/x - 2
= [2(1 + lnx) - 2x]/x
= 2(1 + lnx - x)/x
f est définie sur ]0;+∞[ ⇒ x > 0
⇒ f' est du signe de (1 + lnx - x)
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Thanks 1
siazzle
Merci beaucoup !
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Verified answer
Bonjour,(1 + lnx)² de la forme uⁿ donc dérivée : nu'uⁿ⁻¹ soit 2(1 + lnx)/x
⇒ f'(x) = 2(1 + lnx)/x - 2
= [2(1 + lnx) - 2x]/x
= 2(1 + lnx - x)/x
f est définie sur ]0;+∞[ ⇒ x > 0
⇒ f' est du signe de (1 + lnx - x)