Question

Utilizando a regra de derivação, calcule o ANEXO

Answer 1

$\bullet~~$ Possibilidade 1. Assumindo que a função a ser derivada é $y=\dfrac{\mathrm{\ell n}(x^2)}{\cos 2x}:$

Vamos derivar em relação a $x$ usando a Regra do Quociente combinada com a Regra da Cadeia:

$\dfrac{dy}{dx}=\dfrac{\frac{d}{dx}\!\left[\mathrm{\ell n}(x^2)\right ]\cdot \cos 2x-\mathrm{\ell n}(x^2)\cdot \frac{d}{dx}(\cos 2x)}{(\cos 2x)^2}\\\\\\ \dfrac{dy}{dx}=\dfrac{\left[\frac{1}{x^2}\cdot \frac{d}{dx}(x^2)\right ]\cdot \cos 2x-\mathrm{\ell n}(x^2)\cdot [-\mathrm{sen\,}2x\cdot \frac{d}{dx}(2x)]}{\cos^2 2x}\\\\\\ \dfrac{dy}{dx}=\dfrac{\left[\frac{1}{x^2}\cdot 2x\right ]\cdot \cos 2x-\mathrm{\ell n}(x^2)\cdot [-2\,\mathrm{sen\,}2x]}{\cos^2 2x}\\\\\\ \dfrac{dy}{dx}=\dfrac{\frac{2}{x}\cdot \cos 2x+2\,\mathrm{\ell n}(x^2)\cdot \mathrm{sen\,}2x}{\cos^2 2x}\\\\\\ \dfrac{dy}{dx}=\dfrac{\frac{2}{x}\cdot \cos 2x+2\,\mathrm{\ell n}(x^2)\cdot \mathrm{sen\,}2x}{\cos^2 2x}\cdot \dfrac{x}{x}\\\\\\ \therefore~~\boxed{\begin{array}{c} \dfrac{dy}{dx}=\dfrac{2\cos 2x+2x\,\mathrm{\ell n}(x^2)\cdot \mathrm{sen\,}2x}{x\cos^2 2x} \end{array}}$

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$\bullet~~$ Possibilidade 2. Assumindo que a função a ser derivada é $y=\dfrac{(\mathrm{\ell n}\,x)^2}{\cos 2x}:$

$\dfrac{dy}{dx}=\dfrac{\frac{d}{dx}\!\left[(\mathrm{\ell n\,}x)^2 \right ]\cdot \cos 2x-\mathrm{(\ell n\,}x)^2\cdot \frac{d}{dx}(\cos 2x)}{(\cos 2x)^2}\\\\\\ \dfrac{dy}{dx}=\dfrac{\left[2\,\mathrm{\ell n\,}x\cdot \frac{d}{dx}(\mathrm{\ell n\,}x) \right]\cdot \cos 2x-\mathrm{\ell n^2\,}x\cdot \left[-\mathrm{sen\,} 2x\cdot \frac{d}{dx}(2x) \right ]}{\cos^2 2x}\\\\\\ \dfrac{dy}{dx}=\dfrac{\left[2\,\mathrm{\ell n\,}x\cdot \frac{1}{x} \right]\cdot \cos 2x-\mathrm{\ell n^2\,}x\cdot \left[-2\,\mathrm{sen\,} 2x \right ]}{\cos^2 2x}\\\\\\ \dfrac{dy}{dx}=\dfrac{2\,\mathrm{\ell n}(x)\cdot \frac{1}{x}\cdot \cos 2x+2\,\mathrm{\ell n^2\,}x\cdot \mathrm{sen\,} 2x}{\cos^2 2x}\\\\\\ \dfrac{dy}{dx}=\dfrac{2\,\mathrm{\ell n}(x)\cdot \frac{1}{x}\cdot \cos 2x+2\,\mathrm{\ell n^2\,}x\cdot \mathrm{sen\,} 2x}{\cos^2 2x}\cdot \dfrac{x}{x}$

$\therefore~~\boxed{\begin{array}{c} \dfrac{dy}{dx}=\dfrac{2\,\mathrm{\ell n}(x)\cdot \cos 2x+2x\,\mathrm{\ell n^2\,}x\cdot \mathrm{sen\,} 2x}{x\cos^2 2x} \end{array}}$