Question

Ache a derivada da função h (t) = sen (2t-1)⁵

Answer 1

A função dada é composta:
$h(t)=\sin^5(2t-1)\ |\ h(t)=(h\circ u\circ v)(t)=h(u(v(t))) \\h(u)=u^5\ \ \ \ \ u(v)=\sin(v)\ \ \ \ v(t)=2t-1$
Regra da cadeia:
f'(g(x))=f'(g(x)).g'(x) ou $\frac{df}{dx} =\frac{df}{dg} \frac{dg}{dx}$
então:
$\frac{dh}{dt}= \frac{dh}{du} \frac{du}{dv} \frac{dv}{dt} \\ \\\frac{dh}{du}=5u^4\\\\ \frac{du}{dv}=cos(v)\\\\ \frac{dv}{dt}=2$
substituindo os valores encontrados:

$\\\frac{dh}{dt}=5(u)^4.cos(v).2\ |\ u=sen(v)\ \ \ v=2t-1\implies\\ \frac{dh}{dt}=5\sin^4(2t-1).\cos(2t-1).2=\underline{10\sin^4(2t-1)\cos(2t-1)}=h'(x)\\ obs:(h'=\dot{h}=\frac{dh}{dx})$