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Question Number 180839 by Mastermind last updated on 17/Nov/22

$$\mathrm{Find}\mathrm{the}\mathrm{derivatives}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)\mathrm{of}\mathrm{the}$$$$\mathrm{following}\mathrm{function}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x}:$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{Sin}({\pi }^{\mathrm{Sinx}}+{\pi }^{\mathrm{Cosx}}).$$$$$$$$\mathrm{Mastermind}$$

Answered by saboorhalimi last updated on 17/Nov/22

$$\mathbf{Solution}\mathbf{by}:\mathbf{Saboor}\mathbf{Halimi}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{Sin}({\pi }^{\mathrm{sinx}}+{\pi }^{\cos \left(\mathrm{x}\right)}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=({\pi }^{\sin \left(\mathrm{x}\right)}\cos \left(\mathrm{x}\right)\ln \left(\pi \right)-{\pi }^{\mathrm{cosx}}\mathrm{sinxln}\left(\pi \right))\cos ({\pi }^{\sin \left(\mathrm{x}\right)}+{\pi }^{\cos \left(\mathrm{x}\right)})$$$$$$

Commented by Mastermind last updated on 17/Nov/22

$$\mathrm{With}\mathrm{full}\mathrm{details}\mathrm{explanation}\mathrm{pls}$$

Commented by Acem last updated on 17/Nov/22

$${\left(e^x\right)}^{\prime }=x^{\prime }{e}^x\ln e=1\times e^x\times 1=e^x$$$${\left(a^u\right)}^{\prime }=\ln a\times u^{\prime }\times a^{u}herea=\pi$$$${\left(\cos u\right)}^{\prime }=-u^{\prime }\sin u$$$${\left(\sin u\right)}^{\prime }=u^{{}^{\prime }}\cos u$$$$$$$${\left[\sin ({\pi }^{\sin x}+{\pi }^{\cos x})\right]}^{{}^{\prime }}={({\pi }^{\sin x}+{\pi }^{\cos x})}^{\prime }.\cos ({\pi }^{\sin x}+{\pi }^{\cos x})$$$${({\pi }^{\sin x}+{\pi }^{\cos x})}^{\prime }={\left({\pi }^{\sin x}\right)}^{{}^{\prime }}+{\left({\pi }^{\cos x}\right)}^{\prime }{\left(a^u\right)}^{\prime }lookabove$$$$=\ln \pi .\cos x{\pi }^{\sin x}-\ln \pi .\sin x.{\pi }^{\cos x}$$$$f{\left(x\right)}^{\prime }=\ln \pi (\cos x{\pi }^{\sin x}-\sin x.{\pi }^{\cos x})\cos ({\pi }^{\sin x}+{\pi }^{\cos x})$$$$$$

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