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Question Number 140294 by solihin last updated on 06/May/21

Answered by Satyendra last updated on 06/May/21

$$c){\sin }^{-1}{y}^2=\sin h\left(xy\right)$$$$\Rightarrow \frac{2y}{\sqrt{1-y^4}}{y}^{\prime }=\cos h\left(xy\right)(y+{xy}^{\prime })$$$$\Rightarrow \frac{2y}{\sqrt{1-y^4}}{y}^{\prime }=y\cos h\left(xy\right)+{xy}^{\prime }\cos h\left(xy\right)$$$$\Rightarrow \frac{2y}{\sqrt{1-y^4}}{y}^{\prime }-{xy}^{\prime }\cosh \left(xy\right)=y\cosh \left(xy\right)$$$$\Rightarrow y^{\prime }[\frac{2y}{\sqrt{1-y^4}}-x\cosh \left(xy\right)]=y\cosh \left(xy\right)$$$$\Rightarrow y^{\prime }=\frac{y\cosh \left(xy\right)}{\frac{2y}{\sqrt{1-y^4}}-x\cosh \left(xy\right)}$$$$\Rightarrow y^{\prime }=\frac{y\cosh \left(xy\right)\sqrt{1-y^4}}{2y-x\cosh \left(xh\right)\sqrt{1-y^4}}$$$$\Rightarrow \frac{dy}{dx}=\frac{y\cosh \left(xy\right)\sqrt{1-y^4}}{2y-x\cosh \left(xh\right)\sqrt{1-y^4}}$$

Answered by Satyendra last updated on 06/May/21

$$\left(a\right)y={\left({\cos }^{-1}4x^2\right)}^5$$$$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}{\left({\cos }^{-1}4x^2\right)}^5$$$$\Rightarrow \frac{dy}{dx}=5{\left({\cos }^{-1}4x^2\right)}^4\frac{d}{dx}\left({\cos }^{-1}4x^2\right)$$$$\Rightarrow \frac{dy}{dx}=5{\left({\cos }^{-1}4x^2\right)}^4\frac{1}{-\sqrt{1-{\left(4x^2\right)}^2}}\frac{d}{dx}\left(4x^2\right)$$$$\Rightarrow \frac{dy}{dx}=5{\left({\cos }^{-1}4x^2\right)}^4\frac{1}{-\sqrt{1-16x^4}}8x$$$$\Rightarrow \frac{dy}{dx}=-\frac{40x{\left({\cos }^{-1}4x^2\right)}^4}{\sqrt{1-16x^4}}$$

Answered by liberty last updated on 06/May/21

$$\left(\mathrm{a}\right)\mathrm{y}={\left({\cos }^{-1}{4\mathrm{x}}^2\right)}^5$$$$\Rightarrow {\mathrm{y}}^{\frac{1}{5}}={\cos }^{-1}\left({4\mathrm{x}}^2\right)$$$$\Rightarrow {4\mathrm{x}}^2=\cos \left({\mathrm{y}}^{\frac{1}{5}}\right)$$$$\Rightarrow 8\mathrm{x}=\frac{1}{5}{\mathrm{y}}^{-\frac{4}{5}}(-\sin {\mathrm{y}}^{\frac{1}{5}}).\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{40\mathrm{x}}{{\mathrm{y}}^{-\frac{4}{5}}\left(\sin {\mathrm{y}}^{\frac{1}{5}}\right)}$$$$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-40\mathrm{x}\csc \left({\cos }^{-1}\left({4\mathrm{x}}^2\right)\right).{\left({\cos }^{-1}\left({4\mathrm{x}}^2\right)\right)}^4$$

Answered by qaz last updated on 06/May/21

$$\ll c\gg$$$$\frac{2y}{\sqrt{1-y^2}}{y}^{\prime }=\cosh \left(xy\right)[y+{xy}^{\prime }]=y\cosh \left(xy\right)+{xy}^{\prime }\cosh \left(xy\right)$$$$\Rightarrow y^{\prime }=\frac{y\cosh \left(xy\right)}{\frac{2y}{\sqrt{1-y^2}}-x\cosh \left(xy\right)}$$

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