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Question Number 27379 by abdo imad last updated on 05/Jan/18

$$letgivef\left(x\right)={\int }_x^{2x}\frac{dt}{ln(1+t^2)}calculate{f}^{{}^{\prime }}\left(x\right).$$

Commented by abdo imad last updated on 05/Jan/18

$$iff\left(x\right)={\int }_{\alpha \left(x\right)}^{\beta \left(x\right)}u\left(t\right)dtwith\alpha and\beta arearefunctionofx$$$$f^{{}^{\prime }}\left(x\right)={\beta }^{{}^{\prime }\left(x\right)}u\left(\beta \left(x\right)\right)-{\alpha }^{{}^{\prime }}\left(x\right)u\left(\alpha \left(x\right)\right)intherxercise$$$$f^{{}^{\prime }}\left(x\right)=\frac{{\left(2x\right)}^{{}^{\prime }}}{ln(1+4x^2)}-\frac{{\left(x\right)}^{{}^{\prime }}}{ln(1+x^2)}$$$$=\frac{2}{ln(1+4x^2)}-\frac{1}{ln(1+x^2)}.$$

Answered by prakash jain last updated on 05/Jan/18

$$f^{\prime }\left(x\right)=\frac{1}{ln(1+4x^2)}\frac{d}{dx}2x-\frac{1}{\ln (1+x^2)}\frac{d}{dx}x+{\int }_x^{2x}0dt$$$$=\frac{2}{ln(1+4x^2)}-\frac{1}{\ln (1+x^2)}$$

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