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Question Number 143418 by help last updated on 14/Jun/21

Answered by physicstutes last updated on 14/Jun/21

$$3.1y={\cos }^2{x}^2+{(3-\sqrt{x})}^{30}-2^x$$$$\mathrm{Let}{y}_1={\cos }^2{x}^2,y_1={(3-\sqrt{x})}^{30}\mathrm{and}{y}_3=2^x$$$$\Rightarrow \frac{{dy}_1}{dx}=2x\left(2\cos x(-\sin x)\right)=-4x\sin x\cos x$$$$\Rightarrow \frac{{dy}_2}{dx}=-\frac{1}{2}{x}^{-\frac{1}{2}}\left(30\right){(3-\sqrt{x})}^{29}=-\frac{15}{\sqrt{x}}{(3-\sqrt{x})}^{29}$$$$y_3=2^x\Rightarrow \ln y_3=x\ln 2\Rightarrow \frac{1}{y_3}\frac{{dy}_3}{dx}=\ln 2\Rightarrow \frac{{dy}_3}{dx}=2^x\ln 2$$$$\Rightarrow \frac{dy}{dx}=\frac{{dy}_1}{dx}+\frac{{dy}_2}{dx}-\frac{{dy}_3}{dx}$$$$\Rightarrow \frac{dy}{dx}=-2x\sin 2x-\frac{15}{\sqrt{x}}{(3-\sqrt{x})}^{29}-2^x\ln 2$$

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