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

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

Answered by qaz last updated on 10/Feb/22

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

Commented by MJS_new last updated on 10/Feb/22

$$\mathrm{no}$$$$y={(ax+b)}^c\Rightarrow y^{\prime }=ac{(ax+b)}^{c-1}$$

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