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Find the derivative :

$$f(x)=\left(x^{4}+3 x^{2}-2\right)^{5} \quad y=\sin (\tan (2 x)) \quad f(x)=(2 x-3)^{4}\left(x^{2}+x+1\right)^{5} \quad y=\left(\frac{x^{2}+1}{x^{2}-1}\right)^{2}$$

$$y=\frac{x}{\sqrt{x^{2}+2}} \quad f(x)=\sin (\ln (x)) \quad f(x)=\cos \sqrt{\sin \tan \pi x} \quad y=x^{x}$$

$$y=x^{\sin x} \quad f(x)=\sin \left(e^{x}\right)$$

$$f(x)=\left(x^{4}+3 x^{2}-2\right)^{5} \rightarrow 5\left(x^{4}+3 x^{2}-2\right)^{4} \frac{d}{d x}\left(x^{4}+3 x^{2}-2\right)$$

$$f^{\prime}(x)=5\left(x^{4}+3 x^{2}-2\right)^{4}\left(4 x^{3}+6 x\right)$$

$$y=\sin (\tan (2 x))=\cos (\tan 2 x) \cdot \frac{d}{d x}\left(\tan (2 x)=2 \cos (\tan 2 x) \cdot \sec ^{2}(x)\right.$$

$$f^{\prime}(x)=4(2 x-3) \cdot(2)\left(\left(x^{2}+x+1\right)^{5}\right)+5\left(x^{2}+x+1\right) \cdot(2 x+1)(2 x-3)^{4}$$

$$y=\left(\frac{x^{2}+1}{x^{2}-1}\right)^{2} \rightarrow y^{\prime}=2\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot \frac{d}{d x}\left(\frac{x^{2}+1}{x^{2}-1}\right)$$

$$y^{\prime}=2\left(\frac{x^{2}+1}{x^{2}-1}\right)\left(\frac{(2 x)\left(x^{2}-1\right)-(2 x)\left(x^{2}+1\right)}{\left(x^{2}-1\right)^{2}}\right)$$

$$=\frac{\left(2 x^{2}+2\right)\left(2 x^{3}-2 x-2 x^{3}-2 x\right)}{\left(x^{2}-1\right)^{3}}$$

$$y^{\prime}=\frac{4 x^{5}-4 x^{3}-4 x^{5}-4 x^{3}+4 x^{3}-4 x-4 x^{3}-4 x}{\left(x^{2}-1\right)^{3}}=\frac{-8 x^{3}-8 x}{\left(x^{2}-1\right)^{3}}$$

$$y=\frac{x}{\sqrt{x^{2}+2}}=\frac{(1)\left(\sqrt{x^{2}+2}\right)-\left(\frac{2 x}{2 \sqrt{x^{2}+2}}\right)(x)}{\left(\sqrt{x^{2}+2}\right)^{2}}$$

$$y\prime=\frac{\sqrt{x^{2}+2}-\frac{x^{2}}{\sqrt{x^{2}+2}}}{x^{2}+2}$$

$$=\frac{x^{2}+2-x^{2} / \sqrt{x^{2}+2}}{x^{2}+2}$$

$$y^{\prime}=\frac{2}{\left(x^{2}+2\right) \sqrt{x^{2}+2}}$$

$$f(x)=\sin (\ln (x))=\cos (\ln (x)) \cdot \frac{d}{d x}(\ln (x))$$

$$f^{\prime}(x)=\frac{\cos (\ln (x))}{x}$$

$$f(x)=\cos (\sqrt{\sin \tan \pi x})=-\sin (\sqrt{\sin \tan \pi x}) \cdot \frac{d}{d x} \sqrt{\sin \tan \pi x}$$

$$f^{\prime}(x)=-\sin \sqrt{\sin \tan \pi x} \cdot \frac{\cos (\tan \pi x) \cdot \sec ^{2} \pi x \cdot \pi}{2 \sqrt{\sin \tan \pi x}}$$

$$y=x^{x} \longrightarrow y^{\prime}=\frac{d}{d x}\left(x^{x}\right)$$

$$\ln y=\ln x^{x}$$

$$\ln y=x \ln (x) \longrightarrow \frac{y^{\prime}}{y}=\ln (x)+\frac{1}{x}(x)$$

$$\rightarrow y^{\prime}=y(\ln (x)+1)$$

$$y^{\prime}=y(\ln (x)+1)=x^{x}(\ln (x)+1)$$

$$y=x^{\sin (x)}$$

$$\ln y=\ln x^{\sin (x)} \rightarrow \ln (y)=\sin x \ln (x) \rightarrow \frac{y^{\prime}}{y}=\cos x \ln (x)+\frac{1}{x} \sin (x)$$

$$y^{\prime}=y\left(\cos x \ln x+\frac{1}{x} \sin x\right) \longrightarrow$$

$$y^{\prime}=x^{\sin x}(\cos x \ln (x)+1 / x \sin (x))$$

$$f(x)=\sin \left(e^{x}\right)=\cos \left(e^{x}\right) \cdot \frac{d}{d x}\left(e^{x}\right)=\cos \left(e^{x}\right) \cdot e^{x}$$

$$f^{\prime}(x)=e^{x} \cos e^{x}$$

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