Menu Close

Question-140294

Tinku Tara 0 Comments
Pin




Question Number 140294 by solihin last updated on 06/May/21
Answered by Satyendra last updated on 06/May/21
c) sin^(−1) y^2 =sinh(xy) ⇒((2y)/( (√(1−y^4 )))) y′=cosh(xy) (y+xy′) ⇒((2y)/( (√(1−y^4 )))) y′=ycosh(xy)+xy′cosh(xy) ⇒((2y)/( (√(1−y^4 )))) y′−xy′cosh (xy)=ycosh (xy) ⇒y′[((2y)/( (√(1−y^4 ))))−xcosh (xy)]=ycosh (xy) ⇒y′=((ycosh (xy))/(((2y)/( (√(1−y^4 ))))−xcosh (xy))) ⇒y′=((ycosh (xy)(√(1−y^4 )))/(2y−xcosh (xh)(√(1−y^4 )))) ⇒(dy/dx)=((ycosh (xy)(√(1−y^4 )))/(2y−xcosh (xh)(√(1−y^4 ))))
$$\left.{c}\right)\:\mathrm{sin}^{−\mathrm{1}} {y}^{\mathrm{2}} =\mathrm{sin}{h}\left({xy}\right) \\ $$$$\Rightarrow\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }}\:{y}'=\mathrm{cos}{h}\left({xy}\right) \left({y}+{xy}'\right) \\ $$$$\Rightarrow\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }}\:{y}'={y}\mathrm{cos}{h}\left({xy}\right)+{xy}'\mathrm{cos}{h}\left({xy}\right) \\ $$$$\Rightarrow\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }}\:{y}'−{xy}'\mathrm{cosh}\:\left({xy}\right)={y}\mathrm{cosh}\:\left({xy}\right) \\ $$$$\Rightarrow{y}'\left[\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }}−{x}\mathrm{cosh}\:\left({xy}\right)\right]={y}\mathrm{cosh}\:\left({xy}\right) \\ $$$$\Rightarrow{y}'=\frac{{y}\mathrm{cosh}\:\left({xy}\right)}{\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }}−{x}\mathrm{cosh}\:\left({xy}\right)} \\ $$$$\Rightarrow{y}'=\frac{{y}\mathrm{cosh}\:\left({xy}\right)\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }}{\mathrm{2}{y}−{x}\mathrm{cosh}\:\left({xh}\right)\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{y}\mathrm{cosh}\:\left({xy}\right)\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }}{\mathrm{2}{y}−{x}\mathrm{cosh}\:\left({xh}\right)\sqrt{\mathrm{1}−{y}^{\mathrm{4}} }} \\ $$
Answered by Satyendra last updated on 06/May/21
(a) y=(cos^(−1) 4x^2 )^5 ⇒(dy/dx)=(d/dx)(cos^(−1) 4x^2 )^5 ⇒(dy/dx)=5 (cos^(−1) 4x^2 )^4 (d/(dx ))(cos^(−1) 4x^2 ) ⇒(dy/dx)=5 (cos^(−1) 4x^2 )^4 (1/( −(√(1−(4x^2 )^2 )))) (d/dx)(4x^2 ) ⇒(dy/dx)=5 (cos^(−1) 4x^2 )^4 (1/(−(√(1−16x^(4 ) )))) 8x ⇒(dy/dx)=−((40x(cos^(−1) 4x^2 )^4 )/( (√(1−16x^4 ))))
$$\left({a}\right)\:{y}=\left(\mathrm{cos}^{−\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{5}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{d}}{{dx}}\left(\mathrm{cos}^{−\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{5}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\mathrm{5} \left(\mathrm{cos}^{−\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{4}} \:\:\frac{{d}}{{dx}\:}\left(\mathrm{cos}^{−\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\mathrm{5} \left(\mathrm{cos}^{−\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{4}} \:\frac{\mathrm{1}}{\:−\sqrt{\mathrm{1}−\left(\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }}\:\frac{{d}}{{dx}}\left(\mathrm{4}{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\mathrm{5} \left(\mathrm{cos}^{−\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{4}} \:\frac{\mathrm{1}}{−\sqrt{\mathrm{1}−\mathrm{16}{x}^{\mathrm{4}\:} }}\: \mathrm{8}{x} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=−\frac{\mathrm{40}{x}\left(\mathrm{cos}^{−\mathrm{1}} \mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{4}} }{\:\sqrt{\mathrm{1}−\mathrm{16}{x}^{\mathrm{4}} }} \\ $$
Answered by liberty last updated on 06/May/21
(a) y=(cos^(−1) 4x^2 )^5 ⇒y^(1/5) = cos^(−1) (4x^2 ) ⇒ 4x^2 = cos (y^(1/5) ) ⇒8x = (1/5)y^(−(4/5)) (−sin y^(1/5) ).(dy/dx) ⇒(dy/dx) =− ((40x)/(y^(−(4/5)) (sin y^(1/5) ))) ⇒(dy/dx) = −40x csc (cos^(−1) (4x^2 )).(cos^(−1) (4x^2 ))^4
$$\left(\mathrm{a}\right)\:\mathrm{y}=\left(\mathrm{cos}^{−\mathrm{1}} \mathrm{4x}^{\mathrm{2}} \right)^{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{5}}} \:=\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{4x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\mathrm{4x}^{\mathrm{2}} \:=\:\mathrm{cos}\:\left(\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{5}}} \right) \\ $$$$\Rightarrow\mathrm{8x}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\mathrm{y}^{−\frac{\mathrm{4}}{\mathrm{5}}} \left(−\mathrm{sin}\:\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{5}}} \right).\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=−\:\frac{\mathrm{40x}}{\mathrm{y}^{−\frac{\mathrm{4}}{\mathrm{5}}} \left(\mathrm{sin}\:\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{5}}} \right)} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:−\mathrm{40x}\:\mathrm{csc}\:\left(\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{4x}^{\mathrm{2}} \right)\right).\left(\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{4x}^{\mathrm{2}} \right)\right)^{\mathrm{4}} \\ $$
Answered by qaz last updated on 06/May/21
≪c≫ ((2y)/( (√(1−y^2 ))))y′=cosh (xy)[y+xy′]=ycosh (xy)+xy′cosh (xy) ⇒y′=((ycosh (xy))/(((2y)/( (√(1−y^2 ))))−xcosh (xy)))
$$\ll{c}\gg \\ $$$$\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{y}'=\mathrm{cosh}\:\left({xy}\right)\left[{y}+{xy}'\right]={y}\mathrm{cosh}\:\left({xy}\right)+{xy}'\mathrm{cosh}\:\left({xy}\right) \\ $$$$\Rightarrow{y}'=\frac{{y}\mathrm{cosh}\:\left({xy}\right)}{\frac{\mathrm{2}{y}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}−{x}\mathrm{cosh}\:\left({xy}\right)} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *