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

Find the derivatives f^′ (x) of the following function with respect to x: f(x)=Sin(π^(Sinx) +π^(Cosx) ). Mastermind

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

Answered by saboorhalimi last updated on 17/Nov/22

Solution by: Saboor Halimi f(x)=Sin(π^(sinx) +π^(cos(x)) f ′ (x)= (π^(sin(x)) cos(x)ln(π)−π^(cosx) sinxln(π))cos(π^(sin(x)) +π^(cos(x)) )

$$\:\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{by}}:\:\:\boldsymbol{\mathrm{Saboor}}\:\boldsymbol{\mathrm{Halimi}} \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{Sin}\left(\pi^{\mathrm{sinx}} +\pi^{\mathrm{cos}\left(\mathrm{x}\right)} \right. \\ $$$$\:\mathrm{f}\:'\:\left(\mathrm{x}\right)=\:\left(\pi^{\mathrm{sin}\left(\mathrm{x}\right)} \mathrm{cos}\left(\mathrm{x}\right)\mathrm{ln}\left(\pi\right)−\pi^{\mathrm{cosx}} \mathrm{sinxln}\left(\pi\right)\right)\mathrm{cos}\left(\pi^{\mathrm{sin}\left(\mathrm{x}\right)} +\pi^{\mathrm{cos}\left(\mathrm{x}\right)} \right) \\ $$$$ \\ $$

Commented by Mastermind last updated on 17/Nov/22

With full details explanation pls

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

Commented by Acem last updated on 17/Nov/22

(e^x )′ = x′ e^x ln e= 1×e^x ×1= e^x (a^u )′= ln a× u′×a^u here a= π (cos u)′ = −u′ sin u (sin u)′ = u^′ cos u [sin (π^(sin x) + π^(cos x) )]^′ = (π^(sin x) + π^(cos x) )′ .cos (π^(sin x) + π^(cos x) ) (π^(sin x) + π^(cos x) )′ = (π^(sin x) )^′ + (π^(cos x) )′ (a^u )′ look above = ln π.cos x π^( sin x) − ln π . sin x . π^( cos x) f(x)′= ln π(cos x π^( sin x) − sin x . π^( cos x) )cos (π^(sin x) + π^(cos x) )

$$\left({e}^{{x}} \right)'\:=\:{x}'\:{e}^{{x}} \:\mathrm{ln}\:{e}=\:\mathrm{1}×{e}^{{x}} \:×\mathrm{1}=\:{e}^{{x}} \\ $$$$\:\left({a}^{{u}} \right)'=\:\mathrm{ln}\:{a}×\:{u}'×{a}^{{u}} \:\:\:\:{here}\:{a}=\:\pi \\ $$$$\:\left(\mathrm{cos}\:{u}\right)'\:=\:−{u}'\:\mathrm{sin}\:{u} \\ $$$$\:\left(\mathrm{sin}\:{u}\right)'\:=\:{u}^{'} \:\mathrm{cos}\:{u} \\ $$$$\: \\ $$$$\:\left[\mathrm{sin}\:\left(\pi^{\mathrm{sin}\:{x}} +\:\pi^{\mathrm{cos}\:{x}} \right)\right]^{'} \:=\:\left(\pi^{\mathrm{sin}\:{x}} +\:\pi^{\mathrm{cos}\:{x}} \right)'\:.\mathrm{cos}\:\left(\pi^{\mathrm{sin}\:{x}} +\:\pi^{\mathrm{cos}\:{x}} \right) \\ $$$$\left(\pi^{\mathrm{sin}\:{x}} +\:\pi^{\mathrm{cos}\:{x}} \right)'\:=\:\left(\pi^{\mathrm{sin}\:{x}} \right)^{'} \:+\:\left(\pi^{\mathrm{cos}\:{x}} \right)'\:\:\:\:\:\:\:\:\left({a}^{{u}} \right)'\:{look}\:{above} \\ $$$$\:=\:\mathrm{ln}\:\pi.\mathrm{cos}\:{x}\:\pi^{\:\mathrm{sin}\:{x}} \:−\:\mathrm{ln}\:\pi\:.\:\mathrm{sin}\:{x}\:.\:\pi^{\:\mathrm{cos}\:{x}} \\ $$$${f}\left({x}\right)'=\:\mathrm{ln}\:\pi\left(\mathrm{cos}\:{x}\:\pi^{\:\mathrm{sin}\:{x}} \:−\:\:\mathrm{sin}\:{x}\:.\:\pi^{\:\mathrm{cos}\:{x}} \right)\mathrm{cos}\:\left(\pi^{\mathrm{sin}\:{x}} +\:\pi^{\mathrm{cos}\:{x}} \right) \\ $$$$ \\ $$

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