Question

2) Calcule o coeficiente angular m da reta tangente ao grafico da função f, no ponto de abcissa dado:

c) f(x) = tgx , x = $\pi$

d) f(x) = sec x , x=-$\pi$

Answer 1

O coeficiente angular da reta tangente ao gráfico de uma função é igual o valor da derivada da função calculada naquele ponto.

 
a) $\mathsf{f(x)=tg\,x,}$     no ponto $\mathsf{x=\pi.}$

Derivando,

$\mathsf{f'(x)=(tg\,x)'}\\\\ \mathsf{=\underset{h\to 0}{\ell im}~\dfrac{tg(x+h)-tg\,x}{h}}\\\\\\ \mathsf{=\underset{h\to 0}{\ell im}~\dfrac{\frac{sen(x+h)}{cos(x+h)}-\frac{sin\,x}{cos\,x}}{h}}\\\\\\ \mathsf{=\underset{h\to 0}{\ell im}~\left[\dfrac{sen(x+h)}{cos(x+h)}-\dfrac{sin\,x}{cos\,x} \right]\cdot \dfrac{1}{h}}$

$\mathsf{=\underset{h\to 0}{\ell im}~\left[\dfrac{sen(x+h)\cdot cos\,x}{cos(x+h)\cdot cos\,x}-\dfrac{cos\,(x+h)\cdot sin\,x}{cos\,(x+h)\cdot cos\,x} \right]\cdot \dfrac{1}{h}}\\\\\\ \mathsf{=\underset{h\to 0}{\ell im}~\dfrac{sen(x+h)\cdot cos\,x-cos\,(x+h)\cdot sin\,x}{cos\,(x+h)\cdot cos\,x}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{=\underset{h\to 0}{\ell im}~\dfrac{sen\big[(x+h)-x\big]}{cos\,(x+h)\cdot cos\,x}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{=\underset{h\to 0}{\ell im}~\dfrac{sen\,h}{cos\,(x+h)\cdot cos\,x}\cdot \dfrac{1}{h}}\\\\\\ \mathsf{=\underset{h\to 0}{\ell im}~\dfrac{1}{cos\,(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{h}}$

$\mathsf{=\underset{h\to 0}{\ell im}~\dfrac{1}{cos\,(x+h)\cdot cos\,x}\cdot \underset{h\to 0}{\ell im}~\dfrac{sen\,h}{h}}\\\\\\ \mathsf{=\dfrac{1}{cos\,(x+0)\cdot cos\,x}\cdot 1}\\\\\\ \mathsf{=\dfrac{1}{cos^2\,x}}\\\\\\ \therefore~~\mathsf{f'(x)=sec^2\,x}\qquad\checkmark$


O coeficiente angular no ponto procurado é

$\mathsf{m=f'(\pi)}\\\\ \mathsf{m=sec^2\,\pi}\\\\ \mathsf{m=(-1)^2}\\\\ \boxed{\begin{array}{c}\mathsf{m=1} \end{array}}\qquad\checkmark$

__________

b) $\mathsf{f(x)=sec\,x,}$     no ponto $\mathsf{x=-\pi.}$

Derivando,

$\mathsf{f'(x)=(sec\,x)'}\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\dfrac{sec(x+h)-sec\,x}{h}}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\dfrac{\frac{1}{cos(x+h)}-\frac{1}{cos\,x}}{h}}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{1}{cos(x+h)}-\dfrac{1}{cos\,x}\right]\cdot \dfrac{1}{h}}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x}{cos(x+h)\cdot cos\,x}-\dfrac{cos(x+h)}{cos(x+h)\cdot cos\,x}\right]\cdot \dfrac{1}{h}}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x-cos(x+h)}{cos(x+h)\cdot cos\,x}\right]\cdot \dfrac{1}{h}}$

$=\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x-\big[cos\,x\,cos\,h-sen\,x\,sen\,h\big]}{cos(x+h)\cdot cos\,x}\right]\cdot \dfrac{1}{h}}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x-cos\,x\,cos\,h+sen\,x\,sen\,h}{cos(x+h)\cdot cos\,x}\right]\cdot \dfrac{1}{h}}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x\cdot (1-cos\,h)+sen\,x\,sen\,h}{cos(x+h)\cdot cos\,x}\right]\cdot \dfrac{1}{h}}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{1-cos\,h}{h}+\dfrac{sen\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{h}\right]}$

$=\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{(1-cos\,h)\cdot (1+cos\,h)}{h\cdot (1+cos\,h)}+\dfrac{sen\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{h}\right]}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{1-cos^2\,h}{h\cdot (1+cos\,h)}+\dfrac{sen\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{h}\right]}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen^2\,h}{h\cdot (1+cos\,h)}+\dfrac{sen\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{h}\right]}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{1+cos\,h}\cdot \dfrac{sen\,h}{h}+\dfrac{sen\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{h}\right]}$

$=\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{1+cos\,h} +\dfrac{sen\,x}{cos(x+h)\cdot cos\,x}\right]\cdot \dfrac{sen\,h}{h}}\\\\\\ =\mathsf{\underset{h\to 0}{\ell im}~\left[\dfrac{cos\,x}{cos(x+h)\cdot cos\,x}\cdot \dfrac{sen\,h}{1+cos\,h} +\dfrac{sen\,x}{cos(x+h)\cdot cos\,x}\right]\cdot \underset{h\to 0}{\ell im}~\dfrac{sen\,h}{h}}\\\\\\ =\mathsf{\left[\dfrac{cos\,x}{cos(x+0)\cdot cos\,x}\cdot \dfrac{sen\,0}{1+cos\,0} +\dfrac{sen\,x}{cos(x+0)\cdot cos\,x}\right]\cdot 1}\\\\\\ =\mathsf{0+\dfrac{sen\,x}{cos^2\,x}}$

$=\mathsf{0+\dfrac{sen\,x}{cos\,x}\cdot \dfrac{1}{cos\,x}}\\\\\\ \therefore~~\mathsf{f'(x)=tg\,x\cdot sec\,x}\qquad\checkmark$


O coeficiente angular no ponto procurado é

$\mathsf{m=f'(-\pi)}\\\\ \mathsf{m=tg(-\pi)\cdot sec(-\pi)}\\\\ \mathsf{m=0\cdot (-1)}\\\\ \boxed{\begin{array}{c}\mathsf{m=0} \end{array}}\qquad\checkmark$


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Tags: derivada definição limite função trigonométrica tangente tan tg secante sec coeficiente angular reta tangente ponto cálculo diferencial