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Question Number 165950 by mathlove last updated on 10/Feb/22

y=(2x−1)^(100)   (d^(99) y/dx^(99) )=?∧(d^(85) y/dx^(85) )=?

$${y}=\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{100}} \\ $$$$\frac{{d}^{\mathrm{99}} {y}}{{dx}^{\mathrm{99}} }=?\wedge\frac{{d}^{\mathrm{85}} {y}}{{dx}^{\mathrm{85}} }=? \\ $$

Answered by qaz last updated on 10/Feb/22

y^((1)) =2(2x−1)^(99)   y^((2)) =2^2 (2x−1)^(98)   ...  y^((99)) =2^(99) (2x−1)  y^((85)) =2^(85) (2x−1)^(15)

$$\mathrm{y}^{\left(\mathrm{1}\right)} =\mathrm{2}\left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{99}} \\ $$$$\mathrm{y}^{\left(\mathrm{2}\right)} =\mathrm{2}^{\mathrm{2}} \left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{98}} \\ $$$$... \\ $$$$\mathrm{y}^{\left(\mathrm{99}\right)} =\mathrm{2}^{\mathrm{99}} \left(\mathrm{2x}−\mathrm{1}\right) \\ $$$$\mathrm{y}^{\left(\mathrm{85}\right)} =\mathrm{2}^{\mathrm{85}} \left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{15}} \\ $$

Commented by MJS_new last updated on 10/Feb/22

no  y=(ax+b)^c  ⇒ y′=ac(ax+b)^(c−1)

$$\mathrm{no} \\ $$$${y}=\left({ax}+{b}\right)^{{c}} \:\Rightarrow\:{y}'={ac}\left({ax}+{b}\right)^{{c}−\mathrm{1}} \\ $$

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