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Derivate

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Ricevo da Carmen la seguente domanda:

 

Salve professore,

potrebbe aiutarmi a calcolare le derivate di queste funzioni composte?

\[y=\cot \sqrt{x}\quad y=\ln \left( \ln x \right)\quad y=\frac{\ln \left( {{e}^{x}} \right)}{2{{x}^{4}}}\quad y=\frac{1}{2}{{e}^{2x}}\cos x\quad y=\frac{\ln x}{x-1}\]

\[y={{e}^{\sin x+\cos x}}\quad y=\frac{\ln \left( {{x}^{2}}+1 \right)}{\sqrt{x}}\quad y=\sqrt{\ln \left( {{x}^{2}}+4 \right)}\quad y=\ln \left( {{e}^{x}}-2 \right)\quad y=\ln \left( \tan \frac{x}{2} \right)\quad .\]

Grazie.

 

Le rispondo così:

 

Cara Carmen,

ecco, in ordine, gli svolgimenti:

\[y=\cot \sqrt{x}\to y'=-\frac{1}{{{\sin }^{2}}\sqrt{x}}\cdot \frac{1}{2\sqrt{x}}=-\frac{1}{2\sqrt{x}{{\sin }^{2}}\sqrt{x}}\]

\[y=\ln \left( \ln x \right)\to y'=\frac{1}{\ln x}\cdot \frac{1}{x}=\frac{1}{x\ln x}\]

\[y=\frac{\ln \left( {{e}^{x}} \right)}{2{{x}^{4}}}=\frac{1}{2{{x}^{3}}}=\frac{1}{2}{{x}^{-3}}\to y'=-\frac{3}{2}{{x}^{-4}}=-\frac{3}{2{{x}^{4}}}\]

\[y=\frac{1}{2}{{e}^{2x}}\cos x\to y'={{e}^{2x}}\cos x-\frac{1}{2}{{e}^{2x}}\sin x=\frac{{{e}^{2x}}\left( 2\cos x-\sin x \right)}{2}\]

\[y=\frac{\ln x}{x-1}\to y'=\frac{\frac{x-1}{x}-\ln x}{{{\left( x-1 \right)}^{2}}}=\frac{x\left( 1-\ln x \right)-1}{x{{\left( x-1 \right)}^{2}}}\]

\[y={{e}^{\sin x+\cos x}}\to y'={{e}^{\sin x+\cos x}}\left( \cos x-\sin x \right)\]

\[y=\frac{\ln \left( {{x}^{2}}+1 \right)}{\sqrt{x}}\to \frac{\frac{2x\sqrt{x}}{{{x}^{2}}+1}-\frac{\ln \left( {{x}^{2}}+1 \right)}{2\sqrt{x}}}{x}=\frac{4{{x}^{2}}-\left( {{x}^{2}}+1 \right)\ln \left( {{x}^{2}}+1 \right)}{2{{x}^{\frac{3}{2}}}\left( {{x}^{2}}+1 \right)}\]

\[y=\sqrt{\ln \left( {{x}^{2}}+4 \right)}\to y'=\frac{1}{2\sqrt{\ln \left( {{x}^{2}}+4 \right)}}\cdot \frac{1}{\left( {{x}^{2}}+4 \right)}\cdot 2x=\frac{x}{\left( {{x}^{2}}+4 \right)\sqrt{\ln \left( {{x}^{2}}+4 \right)}}\]

\[y=\ln \left( {{e}^{x}}-2 \right)\to y'=\frac{{{e}^{x}}}{{{e}^{x}}-2}\]

\[y=\ln \left( \tan \frac{x}{2} \right)\to \frac{1}{\tan \frac{x}{2}}\cdot \frac{1}{{{\cos }^{2}}\frac{x}{2}}\cdot \frac{1}{2}=\frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{1}{\sin x}\]

Massimo Bergamini


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