Question

Derivada de (2x+1)^x ?

Answer 1

$f(x)=(2x+1)^x\\\\ f(x)=\big(e^{\mathrm{\ell n}(2x+1)}\big)^x\\\\ f(x)=e^{x\,\mathrm{\ell n}(2x+1)}\\\\ f(x)=e^{g(x)}$


onde $g(x)=x\,\mathrm{\ell n}(2x+1).$


Podemos enxergar $f(x)$ como uma função composta:

$\left\{ \!\begin{array}{l} f(x)=e^{g(x)}\\\\ g(x)=x\,\mathrm{\ell n}(2x+1) \end{array} \right.$


Então, usamos a Regra da Cadeia para derivar:

$f'(x)=\big(e^{g(x)}\big)'\\\\ f'(x)=e^{g(x)}\cdot g'(x)\\\\ f'(x)=f(x)\cdot g'(x)\\\\ f'(x)=(2x+1)^{x}\cdot \big(x\,\mathrm{\ell n}(2x+1)\big)'\\\\ f'(x)=(2x+1)^{x}\cdot \Big((x)'\cdot \mathrm{\ell n}(2x+1)+x\cdot \big(\mathrm{\ell n}(2x+1)\big)'\Big)\\\\ f'(x)=(2x+1)^{x}\cdot \left(1\cdot \mathrm{\ell n}(2x+1)+x\cdot \dfrac{1}{2x+1}\cdot (2x+1)'\right)\\\\\\ f'(x)=(2x+1)^{x}\cdot \left(\mathrm{\ell n}(2x+1)+x\cdot \dfrac{1}{2x+1}\cdot 2\right)\\\\\\ \therefore~~\boxed{\begin{array}{c}f'(x)=(2x+1)^{x}\cdot \left(\mathrm{\ell n}(2x+1)+\dfrac{2x}{2x+1}\right) \end{array}}$


Bons estudos! :-)