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Question Number 53532 by gunawan last updated on 23/Jan/19

$$\mathrm{The}\mathrm{derivative}\mathrm{of}F\left(x\right)=\underset{x^2}{\stackrel{x^3}{\int }}\frac{1}{\log t}dt$$$$(x>0)\mathrm{is}$$

Commented by Abdo msup. last updated on 23/Jan/19

$$\frac{dF}{dx}\left(x\right)=\frac{{\left(x^3\right)}^{{}^{\prime }}}{log\left(x^3\right)}-\frac{{\left(x^2\right)}^{{}^{\prime }}}{log\left(x^2\right)}=\frac{3x^2}{3log\left(x\right)}-\frac{2x}{2log\left(x\right)}$$$$=\frac{x^2-x}{log\left(x\right)}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

$$\frac{dF\left(x\right)}{dx}={\int }_{x^2}^{x^3}\frac{\partial }{\partial x}\left(\frac{1}{logt}\right)dt+\frac{1}{{logx}^3}\times \frac{d}{dx}\left(x^3\right)-\frac{1}{{logx}^2}\times \frac{d}{dx}\left(x^2\right)$$$$=0+\frac{3x^2}{3logx}-\frac{2x}{2logx}$$$$=\frac{x^2}{logx}-\frac{x}{logx}=\frac{x}{logx}(x-1)$$

Commented by gunawan last updated on 23/Jan/19

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}$$

Commented by malwaan last updated on 23/Jan/19

$$0\mathrm{how}?$$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jan/19

$$partialdifferentiationofF\left(t\right)w.r.t.xisxis0$$

Commented by malwaan last updated on 24/Jan/19

$$\mathrm{Yes}$$$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{sir}$$

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