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Question Number 64355 by turbo msup by abdo last updated on 17/Jul/19

$$letF\left(x\right)={\int }_{u\left(x\right)}^{v(x\{}f(x,t)dt$$$$howtocalculate\frac{dF}{dx}\left(x\right)?$$

Commented by MJS last updated on 17/Jul/19

$$\mathrm{I}\mathrm{believe}\mathrm{since}\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}‘‘{\mathrm{status}}^{″}\mathrm{of}$$$$\mathrm{the}\mathrm{constant}\mathrm{factor}x\mathrm{in}\int f(x,t)dt\mathrm{we}\mathrm{cannot}$$$$\mathrm{give}\mathrm{a}\mathrm{general}\mathrm{formula}\mathrm{other}\mathrm{than}\mathrm{this}\mathrm{one}:$$$$G\left(x\right)=\underset{u\left(x\right)}{\stackrel{v\left(x\right)}{\int }}f\left(t\right)dt=F\left(v\left(x\right)\right)-F\left(u\left(x\right)\right)+C$$$$\frac{dG}{dx}=F^{\prime }\left(v\left(x\right)\right)v^{\prime }\left(x\right)-F^{\prime }\left(u\left(x\right)\right)u^{\prime }\left(x\right)$$$$\mathrm{trying}\mathrm{these}\mathrm{simple}\mathrm{examples}\mathrm{I}\mathrm{could}\mathrm{not}\mathrm{see}$$$$\mathrm{a}\mathrm{general}\mathrm{formula}$$$$f(x,t)=x+t$$$$G\left(x\right)=\frac{v^2}{2}+vx-\frac{u^2}{2}-ux$$$$G^{\prime }\left(x\right)=(v+x)v^{\prime }-(u+x)u^{\prime }$$$$f(x,t)=xt$$$$G\left(x\right)=\frac{x}{2}(v^2-u^2)$$$$G^{\prime }\left(x\right)=x(v^{\prime }v-u^{\prime }u)+\frac{1}{2}(v^2-u^2)$$$$f(x,t)=\frac{x}{t}$$$$G\left(x\right)=x\ln \frac{v}{u}$$$$G^{\prime }\left(x\right)=\ln \frac{v}{u}+x(\frac{v^{\prime }}{v}-\frac{u^{\prime }}{u})$$$$f(x,t)=t^x$$$$G\left(x\right)=\frac{v^{x+1}-u^{x+1}}{x+1}$$$$G^{\prime }\left(x\right)=v^x(v^{\prime }+\frac{v\ln v}{x+1}-\frac{v}{{(x+1)}^2})-u^x(u^{\prime }+\frac{u\ln u}{x+1}-\frac{u}{{(x+1)}^2})$$$$f(x,t)=x^t$$$$G\left(x\right)=\frac{x^v-x^u}{\ln x}$$$$G^{\prime }\left(x\right)=x^{v-1}(v^{\prime }x+\frac{v}{\ln x}-\frac{1}{{\ln }^{2}x})-x^{u-1}(u^{\prime }x+\frac{u}{\ln x}-\frac{1}{{\ln }^{2}x})$$

Commented by mathmax by abdo last updated on 17/Jul/19

$$thankyousirmjsforthishardworkwhenifindaformula$$$$forthisiwillpost...$$

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