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Question Number 131775 by Raxreedoroid last updated on 08/Feb/21

$$\mathrm{Find}\mathrm{the}\mathrm{derivative}\mathrm{of}$$$$g\left(x\right)={\int }_{\tan x}^{x^2}\frac{1}{\sqrt{2+t^4}}dt$$

Answered by bemath last updated on 08/Feb/21

$${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\frac{\mathrm{d}\left({\mathrm{x}}^2\right)}{{(2+{\left({\mathrm{x}}^2\right)}^4)}^{1/2}}-\frac{\mathrm{d}\left(\tan \mathrm{x}\right)}{\sqrt{2+\tan {}^4\mathrm{x}}}$$$${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\frac{2\mathrm{x}}{\sqrt{2+{\mathrm{x}}^8}}-\frac{\sec {}^2\mathrm{x}}{\sqrt{2+\tan {}^4\mathrm{x}}}$$$$\mathrm{Formula}\mathrm{f}\left(\mathrm{x}\right)={\int }_{\mathrm{u}\left(\mathrm{x}\right)}^{\mathrm{v}\left(\mathrm{x}\right)}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{v}\left(\mathrm{x}\right)\right).{\mathrm{v}}^{\prime }\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{u}\left(\mathrm{x}\right)\right).{\mathrm{u}}^{\prime }\left(\mathrm{x}\right)$$

Commented by Raxreedoroid last updated on 08/Feb/21

$$Thankssir$$

Answered by mathmax by abdo last updated on 08/Feb/21

$${\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{2\mathrm{x}}{\sqrt{2+{\mathrm{x}}^8}}-\frac{1+{\tan }^2\mathrm{x}}{\sqrt{2+{\tan }^4\mathrm{x}}}$$

Commented by Raxreedoroid last updated on 08/Feb/21

$$Canyoushowthestepssir$$

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