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Question Number 173834 by mnjuly1970 last updated on 19/Jul/22

Commented by infinityaction last updated on 20/Jul/22

$$\mathit{f}\left(\mathit{x}\right)\mathit{g}\left(\mathit{x}\right)\mathit{h}\left(\mathit{x}\right)=({\mathbf{tan}}^7\mathit{x}+{\mathbf{cot}}^7\mathit{x})({\mathbf{tan}}^8\mathit{x}-{\mathbf{cot}}^8\mathit{x})$$$$\times ({\mathbf{tan}}^9\mathit{x}+{\mathbf{cot}}^9\mathit{x})$$$$\mathit{F}\left(\mathit{x}\right)=({\mathbf{tan}}^{15}\mathit{x}-\mathbf{cot}\mathit{x}+\mathbf{tan}\mathit{x}-{\mathbf{cot}}^{15}\mathit{x})({\mathbf{tan}}^9+{\mathbf{cot}}^9\mathit{x})$$$$F\left(x\right)={\tan }^{24}x-{\tan }^{8}x+{\tan }^{10}x-{\cot }^{6}x$$$$+{\tan }^{6}x-{\cot }^{10}x+{\cot }^{8}x-{\cot }^{24}x$$$$F{}^{\prime }\left(x\right)={24\tan }^{23}x{\sec }^{2}x-{8\tan }^{7}x{\sec }^{2}x$$$$+{10\tan }^{9}x{\sec }^{2}x+{6\cot }^{5}x{\mathrm{cosec}}^{2}x$$$$+{6\tan }^{5}x{\sec }^{2}x+{10\cot }^{9}x{\mathrm{cosec}}^{2}x-$$$${8\cot }^{7}x{\mathrm{cosec}}^{2}x+{24\cot }^{23}x{\mathrm{cosec}}^{2}x$$$$F{}^{\prime }\left(x\right)=48-16+20+12+12+20-16+48$$$$\text{Prime causes double exponent: use braces to clarify}$$

Commented by Tawa11 last updated on 19/Jul/22

$$\mathrm{Great}\mathrm{sirs}$$

Answered by mahdipoor last updated on 19/Jul/22

$$P\left(x\right)=P_{a,\pm }\left(x\right)={tan}^a\left(x\right)\pm {cot}^a\left(x\right)$$$$\frac{dP}{dx}=a({tan}^{a-1}\left(x\right).\frac{1}{{cos}^{2}x}\mp {cot}^{a-1}\left(x\right).\frac{1}{{sin}^{2}x})$$$$\Rightarrow \{\begin{array}{l}P\left(\theta \right)=1\pm 1\\ P^{{}^{\prime }}\left(\theta \right)=2a(1\mp 1)\end{array}get\theta =\pi /4$$$$f\equiv P_{7,+}\Rightarrow f\left(\theta \right)=2D_{x}f\left(\theta \right)=0$$$$g\equiv P_{8,-}\Rightarrow g\left(\theta \right)=0D_{x}g\left(\theta \right)=32$$$$h\equiv P_{7,+}\Rightarrow h\left(\theta \right)=2D_{x}h\left(\theta \right)=0$$$$\frac{dF}{dx}\left(\theta \right)=\frac{d}{dx}\left(fgh\right)\left(\theta \right)=(f^{{}^{\prime }}gh+{fg}^{{}^{\prime }}h+{fgh}^{{}^{\prime }})\left(\theta \right)=$$$$0+2\times 32\times 2+0=128$$

Commented by mnjuly1970 last updated on 19/Jul/22

$$verynicemamnonostad$$

Commented by mahdipoor last updated on 20/Jul/22

$$lotfdaridostad:)$$

Answered by mr W last updated on 20/Jul/22

$$F\left(x\right)=f\left(x\right)g\left(x\right)h\left(x\right)$$$$\ln F\left(x\right)=\ln f\left(x\right)+\ln g\left(x\right)+\ln h\left(x\right)$$$$\frac{F^{\prime }\left(x\right)}{F\left(x\right)}=\frac{f^{\prime }\left(x\right)}{f\left(x\right)}+\frac{g^{\prime }\left(x\right)}{g\left(x\right)}+\frac{h^{\prime }\left(x\right)}{h\left(x\right)}$$$$F^{\prime }\left(x\right)=g\left(x\right)h\left(x\right)f^{\prime }\left(x\right)+f\left(x\right)h\left(x\right)g^{\prime }\left(x\right)+f\left(x\right)g\left(x\right)h^{\prime }\left(x\right)$$$$f\left(\frac{\pi }{4}\right)=h\left(\frac{\pi }{4}\right)=2$$$$g\left(\frac{\pi }{4}\right)=0$$$$F^{\prime }\left(\frac{\pi }{4}\right)=f\left(\frac{\pi }{4}\right)h\left(\frac{\pi }{4}\right)g^{\prime }\left(\frac{\pi }{4}\right)=4\times g^{\prime }\left(\frac{\pi }{4}\right)$$$$=4\times {[\frac{8{\tan }^{7}x}{{\cos }^{2}x}+\frac{8{\cot }^{7}x}{{\sin }^{2}x}]}_{x=\frac{\pi }{4}}$$$$=32\times [\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}]$$$$=32\times 4=128$$

Commented by mnjuly1970 last updated on 20/Jul/22

$$reallynicesolutionmaster$$

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