Question

Qual a equação da reta tangente da função f(x)=cosec(x)/x³?​

Answer 1

 $\displaystyle \sf \text{Equa{\c c\~a}o da reta tangente {\`a} fun{\c c\~a}o }f(x) = \frac{cossec(x)}{x^3 } \text{ no ponto P} (x_o,y_o) : \\\\ y-y_o=m(x-x_o) \\\\ onde : \\\\ m = f'(x_o) \\\\ \text{Derivando a fun{\c c\~a}o} : \\\\ f'(x) = \left(\frac{cossec(x)}{x^3}\right)' \\\\\\ \text{Use a regra do quociente :} \\\\ \left(\frac{f}{g}\right)' = \frac{f'\cdot g-f\cdot g'}{g^2} \\\\\\ Da{\'i}}: \\\\ f'(x) = \frac{[cossec(x)]'\cdot x^3-cossec(x)\cdot [x^3]'}{(x^3)^2}$

$\displaystyle \sf f'(x) = \frac{ -cossec(x).cotg (x)\cdot x^3-cossec(x) \cdot 3x^2}{x^6 } \\\\\\ f'(x) = \frac{x^2\cdot [-cossec(x).cotg (x)\cdot x-3\cdot cossec(x)]}{x^6} \\\\\\ f'(x) = \frac{-cossec(x).cotg (x)\cdot x-3\cdot cossec(x)}{x^4} \\\\\\ \text{substituindo o ponto }x_o : \\\\ m = f'(x_o) = \frac{-cossec(x_o).cotg (x_o)\cdot (x_o)-3\cdot cossec(x_o)}{(x_o)^4} \\\\\\ m = \frac{-cossec(x_o).cotg (x_o)\cdot (x_o)-3\cdot cossec(x_o)}{(x_o)^4}$

Daí a equação da reta tangente à função seria do tipo :

$\displaystyle \sf \boxed{\sf\ y-y_o = \left[ \frac{-cossec(x_o).cotg (x_o)\cdot (x_o)-3\cdot cossec(x_o)}{(x_o)^4} \right]\cdot (x-x_o) \ }$

Se quisesse achar o valor do ponto $\sf y_o$, basta substituir $\sf x_o$ na equação da função :

$\displaystyle \sf y_o = \frac{cossec(x_o)}{(x_o)^3} \\\\\ \text{Da{\'i}, a equa{\c c\~a}o da reta tangente {\`a} fun{\c c}{\~a}o ser{\`a}}: \\\\\ y-y_o = m(x-x_o) \\\\ \boxed{\sf \ y-\frac{cossec(x_o)}{(x_o)^3} =\left[\frac{-cossec(x_o).cotg (x_o)\cdot (x_o)-3\cdot cossec(x_o)}{(x_o)^4} \right](x-x_o ) \ }\checkmark$