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Question Number 102698 by bramlex last updated on 10/Jul/20

Answered by bramlex last updated on 10/Jul/20

$$\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{h}\left(\mathrm{x}\right)\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{h}}^{\prime }\left(\mathrm{x}\right)$$$$5+x^{2}f\left(x\right)=16x^3-\frac{15}{2}\sqrt{x}$$$$x^{2}f\left(x\right)=16x^3-\frac{15\sqrt{x}}{2}-5$$$$f\left(x\right)=16x-\frac{15}{2x\sqrt{x}}-\frac{5}{x^2}=16\mathrm{x}-\frac{15}{2}{\mathrm{x}}^{-3/2}-{5\mathrm{x}}^{-2}$$$$f{}^{\prime }\left(\mathrm{x}\right)=16+\frac{45}{{4\mathrm{x}}^2\sqrt{\mathrm{x}}}+\frac{10}{{\mathrm{x}}^{-3}}$$

Answered by floor(10²Eta[1]) last updated on 10/Jul/20

$$5+{\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)={16\mathrm{x}}^3-\frac{15\sqrt{\mathrm{x}}}{2}$$$$\mathrm{f}\left(\mathrm{x}\right)=16\mathrm{x}-\frac{15}{2}{\mathrm{x}}^{-3/2}-{5\mathrm{x}}^{-2}$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=16+\frac{45}{4}{\mathrm{x}}^{-5/2}+{10\mathrm{x}}^{-3}$$$${\mathrm{f}}^{\prime }(-2)=16+\frac{45}{4}\times \frac{1}{\sqrt{{(-2)}^5}}+10\times \frac{1}{{(-2)}^3}$$$$=\frac{59}{4}+\frac{45}{16\mathrm{i}\sqrt{2}}$$

Commented by bramlex last updated on 10/Jul/20

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