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Question Number 204426 by universe last updated on 17/Feb/24

$$\mathrm{let}\mathrm{a},\mathrm{b}>0\mathrm{find}\mathrm{all}\mathrm{differentiable}\mathrm{function}$$$$\mathrm{f}:(0,\mathrm{\infty })\to (0,\mathrm{\infty })\mathrm{such}\mathrm{that}$$$${\mathrm{f}}^{\prime }\left(\frac{a}{\mathrm{x}}\right)=\frac{\mathrm{bx}}{\mathrm{f}\left(\mathrm{x}\right)},\mathrm{\forall }\mathrm{x}>0$$

Answered by MrGaster last updated on 03/Feb/25

$$\mathrm{Let}f\left(x\right)=\sqrt{{bx}^2+c},c=\frac{a^{2}b}{4}$$$$f^{\prime }\left(x\right)=\frac{bx}{\sqrt{{bx}^2+c}}$$$$\mathrm{Substitute}x=\frac{a}{y}\mathrm{into}{f}^{\prime }\left(x\right):$$$$f^{\prime }\left(\frac{a}{y}\right)=\frac{b\left(\frac{a}{y}\right)}{\sqrt{b{\left(\frac{a}{y}\right)}^2+c}}=\frac{\frac{ba}{y}}{\sqrt{\frac{{ba}^2}{y^2}+c}}$$$$\frac{\frac{ba}{y}}{\sqrt{\frac{{ba}^2}{y^2}+c}}=\frac{by}{f\left(y\right)}$$$$\mathrm{Substitute}f\left(y\right)=\sqrt{{by}^2+c}:$$$$\frac{\frac{by}{y}}{\sqrt{\frac{{ba}^2}{y^2}+c}}=\frac{by}{f\left(y\right)}$$$$\mathrm{Substitute}f\left(y\right)=\sqrt{{by}^2+c}:$$$$\frac{ba}{\frac{y}{\sqrt{\frac{{ba}^2}{y^2}+c}}}=\frac{by}{\sqrt{{by}^2+c}}$$$$\frac{ba}{y}\cdot \sqrt{{by}^2+c}=by\cdot \sqrt{\frac{{ba}^2}{y^2}+c}$$$${\left(\frac{ba}{y}\right)}^2({by}^2+c)=b^2{y}^2(\frac{{ba}^2}{y^2}+c)$$$$\frac{b^2{a}^2}{y^2}({by}^2+c)=b^2{y}^2(\frac{{ba}^2}{y^2}+c)$$$$\frac{a^2}{y^2}({by}^2+c)=y^2(\frac{a^2}{y^2}+c)$$$$a^{2}b+\frac{a^{2}c}{y^2}=a^2+{cy}^2$$$$\mathrm{Subtitute}c=\frac{a^{2}b}{4}:$$$$a^{2}b+\frac{a^2\cdot \frac{a^{2}b}{4}}{y^2}=a^2+\frac{a^{2}b}{4}{y}^2$$$$4a^2{by}^2+a^{4}b=4a^2{y}^2+a^2{by}^2$$$$a^2{by}^4-4a^2{by}^2+a^{4}b=0$$$$y^4-4y^2+a^2=0$$$$\mathrm{Let}z=y^2:$$$$z^2-4z+a^2=0$$$$z=2\pm \sqrt{4-a^2}$$$$y=\sqrt{2\pm \sqrt{4-a^2}}$$$$\mathrm{so}:$$$$f\left(x\right)=\sqrt{{bx}^2+\frac{a^{2}b}{4}}$$

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