Question

Derivadas !! me ajudem por favor...

Calcule as derivadas laterais abaixo nos pontos onde a função dada não é derivável e faça os gráficos. obs: explicar passo a passo da resolução.

a) $y=\begin{pmatrix}-2x&se\:\le 1.\\ 3x-5&se\:\ge 1.\end{pmatrix}$

b) $y=\left|x-1\right|\left(x-3\right).$

c) $y=tan\:\frac{\sqrt{x}}{x+1}$

Answer 1

a)
$y'=-2\ se\ \le1\\ y'=3\ se \ \textgreater \ 1$

b)
$|x-1|= \left \{ {{x-1\ se\ x\ge1} \atop {-x+1\ se\ x\ \textless \ 1}} \right. \\ y=(x-1)(x-3)\ se\ x\ge1\\ y'=(x-1)'.(x-3)+(x-1).(x-3)'=\\ =1.(x-3)+(x-1).1=x-3+x-1=2x-4\ se\ x\ge1\\ \\ y=(-x+1)(x-3)\ se\ x\ \textless \ 1\\ y'=(-x+1)'.(x-3)+(-x+1).(x-3)'=\\ =-1.(x-3)+(-x+1).1=-x+3-x+1=\\ =-2x+4\ se\ x\ \textless \ 1$

c)
$y=\tan\frac{\sqrt x}{x+1}\\ y'=(\tan\frac{\sqrt x}{x+1})'=\frac1{\cos^2\frac{\sqrt x}{x+1}}.(\frac{\sqrt x}{x+1})'=\\ = \frac1{\cos^2\frac{\sqrt x}{x+1}}.\frac{(\sqrt x)'.(x+1)-\sqrt x.(x+1)'}{(x+1)^2}=\\ = \frac1{\cos^2\frac{\sqrt x}{x+1}}.\frac{\frac1{2\sqrt x}.(x+1)-\sqrt x.1}{(x+1)^2}= \frac1{\cos^2\frac{\sqrt x}{x+1}}.\frac{\frac1{2\sqrt x}.(x+1)-\sqrt x}{(x+1)^2}$