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Question Number 186726 by Rupesh123 last updated on 09/Feb/23

Answered by cortano12 last updated on 09/Feb/23

$$f{}^{\prime }\left(x\right)=\frac{3x^2}{1+\ln x^3}=\frac{3x^2}{\ln {ex}^3}$$$$f{}^{″}\left(x\right)=\frac{6x.\ln {ex}^3-\frac{3{ex}^2}{{ex}^3}.3x^2}{{\left(\ln {ex}^3\right)}^2}$$

Answered by Ar Brandon last updated on 09/Feb/23

$$\frac{\partial }{\partial t}{\int }_{u\left(t\right)}^{\mathrm{v}\left(t\right)}f(x,t)dx={\int }_{u\left(t\right)}^{\mathrm{v}\left(t\right)}\frac{\partial }{\partial t}f(x,t)dx+\left[f(\mathrm{v}\left(t\right),t)\right]{\mathrm{v}}^{\prime }\left(t\right)-\left[f(u\left(t\right),t)\right]u^{\prime }\left(t\right)$$

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