1-y-arcsin-sinx-y-2-y-sin-x-1-y-3-y-ln-5-sinx-y-4-y-cos-2x-3-y-

Question Number 169356 by Shrinava last updated on 29/Apr/22

$$\mathrm{1}.\:\mathrm{y}\:=\:\mathrm{arcsin}\left(\mathrm{sinx}\right)\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{2}.\:\mathrm{y}\:=\:\mathrm{sin}\:\sqrt{\mathrm{x}\:+\:\mathrm{1}}\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{3}.\:\mathrm{y}\:=\:\mathrm{ln}^{\mathrm{5}} \:\mathrm{sinx}\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{4}.\:\mathrm{y}\:=\:\mathrm{cos}\left(\mathrm{2x}\:+\:\mathrm{3}\right)\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$

Answered by thfchristopher last updated on 29/Apr/22

1. y=sin^(−1) sin x ⇒y=x ⇒y^′ =1 2. y=sin (√(x+1)) y^′ =(cos (√(x+1)))((1/(2(√(x+1))))) =((cos (√(x+1)))/(2(√(x+1)))) 3. Do you mean y=(ln sin x)^5 ? y^′ =[5(ln sin x)^4 ]((1/(sin x)))(cos x) =5cot x(ln sin x)^4 4. y=cos (2x+3) y^′ =-2sin (2x+3)

$$\mathrm{1}.\:{y}=\mathrm{sin}^{−\mathrm{1}} \mathrm{sin}\:{x} \\ $$$$\Rightarrow{y}={x} \\ $$$$\Rightarrow{y}^{'} =\mathrm{1} \\ $$$$ \\ $$$$\mathrm{2}.\:{y}=\mathrm{sin}\:\sqrt{{x}+\mathrm{1}} \\ $$$${y}^{'} =\left(\mathrm{cos}\:\sqrt{{x}+\mathrm{1}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}\right) \\ $$$$=\frac{\mathrm{cos}\:\sqrt{{x}+\mathrm{1}}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}} \\ $$$$ \\ $$$$\mathrm{3}.\:\mathrm{Do}\:\mathrm{you}\:\mathrm{mean}\:{y}=\left(\mathrm{ln}\:\mathrm{sin}\:{x}\right)^{\mathrm{5}} \:? \\ $$$${y}^{'} =\left[\mathrm{5}\left(\mathrm{ln}\:\mathrm{sin}\:{x}\right)^{\mathrm{4}} \right]\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\right)\left(\mathrm{cos}\:{x}\right) \\ $$$$=\mathrm{5cot}\:{x}\left(\mathrm{ln}\:\mathrm{sin}\:{x}\right)^{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{4}.\:{y}=\mathrm{cos}\:\left(\mathrm{2}{x}+\mathrm{3}\right) \\ $$$${y}^{'} =-\mathrm{2sin}\:\left(\mathrm{2}{x}+\mathrm{3}\right) \\ $$

Answered by Mathspace last updated on 29/Apr/22

1)y=arcsin(sinx) ⇒y^′ =((cosx)/( (√(1−sin^2 x)))) =ξ ×1 =ξ with ξ^2 =1

$$\left.\mathrm{1}\right){y}={arcsin}\left({sinx}\right)\:\Rightarrow{y}^{'} =\frac{{cosx}}{\:\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {x}}} \\ $$$$=\xi\:×\mathrm{1}\:=\xi\:\:{with}\:\xi^{\mathrm{2}} =\mathrm{1} \\ $$

Answered by Mathspace last updated on 29/Apr/22

2) y=sin((√(x+1))) ⇒y^′ =(1/(2(√(x+1))))cos((√(x+1)))

$$\left.\mathrm{2}\right)\:{y}={sin}\left(\sqrt{{x}+\mathrm{1}}\right)\:\Rightarrow{y}^{'} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}{cos}\left(\sqrt{{x}+\mathrm{1}}\right) \\ $$

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