Question

Considere a função f(x)=(ln(x²)+5x³)/1+cos²x, o valor de f'(x)=(π/2) é igual a

Answer 1

• A derivada da função f(x) no ponto π/2 é igual a:

   $\Large\displaystyle\begin{gathered}f'\left( \frac{\pi}{2}\right)=\dfrac{16+15\pi^3}{4\pi}\end{gathered}$

Para resolver sua questão, iremos primeiramente aplicar a derivada do quociente, dada da seguinte forma:

           $\Large\displaystyle\text{ \begin{gathered}\underline{\boxed{ \left( \frac{f}{g}\right)'=\frac{f'\cdot g- f\cdot g'}{g^2}}} \end{gathered} }$

Aplicando na sua questão, temos que:

$\Large\displaystyle\begin{gathered}f(x)=\frac{\ln(x^2)+5x^3}{1+\cos^2(x)} \end{gathered}$

$\large\displaystyle\begin{gathered}f'(x)=\frac{\left(\ln(x^2)+5x^3\right)'\left(1+\cos^2(x)\right)-\left(\ln(x^2)+5x^3\right)\left(1+\cos^2(x)\right)'}{\left(1+\cos^2(x)\right)^2} \end{gathered}$

Vale ressaltar também as seguintes fórmulas:

 $\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\cos^2(x)=\frac{1+\cos(2x)}{2}}} \ \ \wedge \ \ \underline{\boxed{(a+b)^2=a^2+2ab+b^2}}\end{gathered} }$

Sabendo disso, logo:

$\large\displaystyle\text{ \begin{gathered}f'(x)=\frac{\left(\ln(x^2)+5x^3\right)'\left(1+\cos^2(x)\right)-\left(\ln(x^2)+5x^3\right)\left(1+\cos^2(x)\right)'}{\left(1+\dfrac{1+\cos(2x)}{2} \right)^2} \end{gathered} }$

$\large\displaystyle\text{ \begin{gathered}f'(x)=\frac{\left(\ln(x^2)+5x^3\right)'\left(1+\cos^2(x)\right)-\left(\ln(x^2)+5x^3\right)\left(1+\cos^2(x)\right)'}{1^2+\!\diagup\!\!\!\!2\cdot\left( \dfrac{1+\cos(2x)}{\!\diagup\!\!\!\!2} \right)+\left(\dfrac{1+\cos(2x)}{2}\right)^2}\end{gathered} }$

Resolvendo as derivadas, temos que:

$\Large\displaystyle\begin{gathered} \frac{d(\ln(x^2)+5x^3)}{dx} =\frac{2}{x}+15x^2 \end{gathered}$

$\Large\displaystyle\begin{gathered} \frac{d(1+\cos^2x)}{dx} =-\sin(2x) \end{gathered}$

Ficando então:

$\large\displaystyle\text{ \begin{gathered}f'(x)=\frac{\left(\dfrac{2}{x}+15x^2 \right)\left(1+\cos^2(x)\right)+\sin(2x)\left(\ln(x^2)+5x^3\right)}{2+\cos(2x)+\left(\dfrac{1+\cos(2x)}{2}\right)^2}\end{gathered} }$

Aplicando o limite em f'(x) quando x tende a π/2, temos que:

$\large\displaystyle\text{ \begin{gathered} \lim_{x \to \frac{\pi}{2}} \frac{\left(\dfrac{2}{x}+15x^2 \right)\left(1+\cos^2(x)\right)+\sin(2x)\left(\ln(x^2)+5x^3\right)}{2+\cos(2x)+\left(\dfrac{1+\cos(2x)}{2}\right)^2}\end{gathered} }$

$\large\displaystyle\text{ \begin{gathered} \frac{\left(\dfrac{4}{\pi}+15\left(\dfrac{\pi}{2}\right)^2 \right)\left(1+\cos^2\left(\dfrac{\pi}{2}\right)\right)+\sin(\pi)\left(\ln\left(\dfrac{\pi}{2}\right)^2+5\left(\dfrac{\pi}{2}\right)^3\right)}{2+\cos(\pi)+\left(\dfrac{1+\cos(\pi)}{2}\right)^2}\end{gathered} }$

Agora é só simplificar, logo:

$\large\displaystyle\text{ \begin{gathered} f'\left( \frac{\pi}{2}\right)=\frac{\left(\dfrac{4}{\pi}+\dfrac{15\pi^2}{4} \right)+0}{2-1+\left(\dfrac{1-1}{2}\right)^2}\end{gathered} }$

$\large\displaystyle\begin{gathered} f'\left( \frac{\pi}{2}\right)=\dfrac{4}{\pi}+\dfrac{15\pi^2}{4}\end{gathered}$

$\large\displaystyle\text{ \begin{gathered}\therefore \green{\underline{\boxed{ f'\left( \frac{\pi}{2}\right)=\dfrac{16+15\pi^3}{4\pi}}}}\ \ (\checkmark ).\end{gathered} }$

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