Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 143274 by Aditya9886 last updated on 12/Jun/21

Answered by Olaf_Thorendsen last updated on 12/Jun/21

$$y={\sin }^2\left(\cos x\right)$$$$\frac{dy}{dx}=(-\sin x)\times 2\cos \left(\cos x\right)\times \sin \left(\cos x\right)$$$$\frac{dy}{dx}=-\sin x.\sin \left(2\cos x\right)$$$$$$$$\mathrm{Nota}:$$$$y={\sin }^2\left(\cos x\right)=\frac{1-\cos \left(2\cos x\right)}{2}$$$$\frac{dy}{dx}=-\frac{(-\sin x)(-2\sin \left(2\cos x\right))}{2}$$$$\frac{dy}{dx}=-\sin x.\sin \left(2\cos x\right)$$

Answered by justtry last updated on 12/Jun/21

$$u=cosx\Rightarrow \frac{du}{dx}=-sinx$$$$y={sin}^{2}u\Rightarrow \frac{dy}{du}=2sinu.cosu=sin2u=sin2\left(cosx\right)$$$$\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}=sin2\left(cosx\right).(-sinx)=-sinx.sin2\left(cosx\right)$$

Answered by mathmax by abdo last updated on 12/Jun/21

$$\mathrm{we}\mathrm{have}\mathrm{y}={\sin }^2\left(\mathrm{cosx}\right)\Rightarrow \mathrm{y}=\frac{1-\cos \left(2\mathrm{cosx}\right)}{2}\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2}(-1)(-2\mathrm{sinx})\sin \left(2\mathrm{cosx}\right)\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{sinx}\sin \left(2\mathrm{cosx}\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com