Question

Seja f(x)=ln(x2+7/ βx+7) . Ache β ∈ R de modo que f′(1) =2/9 .

Answer 1

Resposta:

$\sf f(x)=ln\bigg(\dfrac{x^2+7}{\beta x+7}\bigg)$

$\sf f'(x)=\dfrac{d}{dx}ln\bigg(\dfrac{x^2+7}{\beta x+7}\bigg)$

Aplicando a regra da cadeia: dy/dx = dy/du . du/dx, faça u = (x² + 7)/(βx + 7):

$\sf f'(x)=\dfrac{d}{du}ln(u)\cdot\dfrac{d}{dx}u$

$\sf f'(x)=\dfrac{1}{u}\cdot\dfrac{d}{dx}u$

$\sf f'(x)=\dfrac{1}{\frac{x^2\:+\:7}{\beta x\:+\:7}}\cdot\dfrac{d}{dx}\bigg(\dfrac{x^2+7}{\beta x+7}\bigg)$

$\sf f'(x)=\dfrac{\beta x+7}{x^2+7}\cdot\dfrac{d}{dx}\bigg(\dfrac{x^2+7}{\beta x+7}\bigg)$

Regra do quociente: d/dx(f/g) = (df/dx . g - dg/dx . f)/g²

$\sf f'(x)=\dfrac{\beta x+7}{x^2+7}\cdot\dfrac{\frac{d}{dx}(x^2+7)\cdot(\beta x+7)-\frac{d}{dx}(\beta x+7)\cdot(x^2+7)}{(\beta x+7)^2}$

$\sf f'(x)=\dfrac{\beta x+7}{x^2+7}\cdot\dfrac{(\frac{d}{dx}x^2+\frac{d}{dx}7)\cdot(\beta x+7)-(\frac{d}{dx}\beta x+\frac{d}{dx}7)\cdot(x^2+7)}{(\beta x+7)^2}$

$\sf f'(x)=\dfrac{\beta x+7}{x^2+7}\cdot\dfrac{(2x+0)\cdot(\beta x+7)-(\beta+0)\cdot(x^2+7)}{(\beta x+7)^2}$

$\sf f'(x)=\dfrac{\beta x+7}{x^2+7}\cdot\dfrac{2x(\beta x+7)-\beta(x^2+7)}{(\beta x+7)^2}$

$\sf f'(x)=\dfrac{\beta x+7}{x^2+7}\cdot\dfrac{2\beta x^2+14x-\beta x^2-7\beta)}{(\beta x+7)^2}$

$\sf f'(x)=\dfrac{(\beta x+7)(2\beta x^2+14x-\beta x^2-7\beta)}{(x^2+7)(\beta x+7)^2}$

Dado que f'(1) = 2/9:

$\sf f'(1)=\dfrac{(\beta \cdot1+7)(2\beta \cdot1^2+14\cdot1-\beta \cdot1^2-7\beta)}{(1^2+7)(\beta \cdot1+7)^2}=\dfrac{2}{9}$

$\sf \dfrac{(\beta+7)(2\beta \cdot1+14-\beta \cdot1-7\beta)}{(1+7)(\beta +7)^2}=\dfrac{2}{9}$

$\sf \dfrac{(\beta+7)(2\beta+14-\beta-7\beta)}{8(\beta +7)^2}=\dfrac{2}{9}$

$\sf \dfrac{(\beta+7)(14-6\beta)}{8(\beta +7)^2}=\dfrac{2}{9}$

$\sf \dfrac{2(\beta+7)(7-3\beta)}{8(\beta +7)^2}=\dfrac{2}{9}$

$\sf \dfrac{7-3\beta}{4(\beta +7)}=\dfrac{2}{9}$

$\sf \dfrac{7-3\beta}{4\beta +28}=\dfrac{2}{9}$

$\sf (4\beta +28)\cdot2=(7-3\beta)\cdot9$

$\sf 8\beta +56=63-27\beta$

$\sf 8\beta+27\beta=63-56$

$\sf 35\beta=7$

$\sf \beta=\dfrac{7}{35}$

$\red{\boxed{\sf \beta=\dfrac{1}{5}}}$