Question

Derivada de Cosx^senx ????

Answer 1

Calcular a derivada da função

     $\large\begin{array}{l}f(x)=\cos x^{\mathrm{sen\,}x}\end{array}$


Essa é uma função composta de exponencial. Então, vamos usar alguns artifícios com logaritmos.

Aplicando logaritmos a ambos os lados, temos

     $\large\begin{array}{l}\ln\!\big[f(x)\big]=\ln\!\big(\!\cos x^{\mathrm{sen\,}x}\big)\end{array}\\\\\\ \large\begin{array}{l}\ln\!\big[f(x)\big]=\mathrm{sen\,}x\cdot \ln(\cos x)\end{array}$


Derive ambos os lados implicitamente. Lembre-se de aplicar a Regra da Cadeia:

     $\large\begin{array}{l}\dfrac{d}{dx}\ln\!\big[f(x)\big]=\dfrac{d}{dx}\big[\mathrm{sen\,}x\cdot \ln(\cos x)\big] \\\\ \dfrac{1}{f(x)}\cdot \dfrac{d}{dx}\big[f(x)\big]=\dfrac{d}{dx}\big[\mathrm{sen\,}x\cdot \ln(\cos x)\big]\\\\ \dfrac{1}{\cos x^{\mathrm{sen\,}x}}\cdot \dfrac{d}{dx}\big[f(x)\big]=\dfrac{d}{dx}(\mathrm{sen}\,x)\cdot \ln(\cos x)+\mathrm{sen\,}x\cdot \dfrac{d}{dx}\big[\ln(\cos x)\big]\\\\ \dfrac{1}{\cos x^{\mathrm{sen\,}x}}\cdot \dfrac{d}{dx}\big[f(x)\big]=\cos x\cdot \ln(\cos x)+\mathrm{sen\,}x\cdot \dfrac{1}{\cos x}\cdot \dfrac{d}{dx}(\cos x)\\\\ \dfrac{1}{\cos x^{\mathrm{sen\,}x}}\cdot \dfrac{d}{dx}\big[f(x)\big]=\cos x\cdot \ln(\cos x)+\mathrm{sen\,}x\cdot \dfrac{1}{\cos x}\cdot (-\,\mathrm{sen\,}x) \end{array}$

     $\large\begin{array}{l} \dfrac{1}{\cos x^{\mathrm{sen\,}x}}\cdot \dfrac{d}{dx}\big[f(x)\big]=\cos x\cdot \ln(\cos x)-\dfrac{\mathrm{sen^2\,}x}{\cos x}\\\\\\ \dfrac{1}{\cos x^{\mathrm{sen\,}x}}\cdot \dfrac{d}{dx}\big[f(x)\big]=\cos x\cdot \ln(\cos x)-\dfrac{1-\cos^2 x}{\cos x}\\\\\\ \dfrac{1}{\cos x^{\mathrm{sen\,}x}}\cdot \dfrac{d}{dx}\big[f(x)\big]=\cos x\cdot \ln(\cos x)-\dfrac{1}{\cos x}+\dfrac{\cos^2 x}{\cos x}\\\\\\ \dfrac{1}{\cos x^{\mathrm{sen\,}x}}\cdot \dfrac{d}{dx}\big[f(x)\big]=\cos x\cdot \ln(\cos x)-\sec x+\cos x \end{array}$


Finalmente chegamos a

     $\large\begin{array}{l} \dfrac{d}{dx}\big[f(x)\big]=\cos x^{\mathrm{sen\,}x}\cdot \big[\cos x\cdot \ln(\cos x)-\sec x+\cos x\big] \end{array}$


Dúvidas? Comente.


Bons estudos! :-)