Question

encontre a derivada mediante a regra da cadeia:
$r(t)= 10^{2\sqrt{t} }$

Answer 1

Explicação passo-a-passo:

$\sf r(t)=10^{2\sqrt{t}}$

Seja $\sf u(t)=2\sqrt{t}$

$\sf r(u)=10^u$

Pela regra da cadeia:

$\sf \dfrac{dr}{dt}=\dfrac{dr}{du}\cdot\dfrac{du}{dt}$

Temos que:

• $\sf r(u)=10^u$

$\sf \dfrac{dr}{du}=ln~(10)\cdot10^{u}$

$\sf \dfrac{dr}{du}=ln~(10)\cdot10^{2\sqrt{t}}$

• $\sf u(t)=2\sqrt{t}$

$\sf u(t)=2\cdot t^{\frac{1}{2}}$

$\sf \dfrac{du}{dt}=\dfrac{1}{2}\cdot2\cdot t^{\frac{1}{2}-1}$

$\sf \dfrac{du}{dt}=\dfrac{2}{2}\cdot t^{\frac{1-2}{2}}$

$\sf \dfrac{du}{dt}=1\cdot t^{\frac{-1}{2}}$

$\sf \dfrac{du}{dt}=t^{\frac{-1}{2}}$

$\sf \dfrac{du}{dt}=\dfrac{1}{\sqrt{t}}$

Assim:

$\sf \dfrac{dr}{dt}=ln~(10)\cdot10^{2\sqrt{t}}\cdot\dfrac{1}{\sqrt{t}}$

$\sf \red{\dfrac{dr}{dt}=\dfrac{ln~(10)\cdot10^{2\sqrt{t}}}{\sqrt{t}}}$

Answer 2

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$\sf r(t)=10^{2\sqrt{t}}\\\sf fac_{\!\!,}a~u=2\sqrt{t}\\\sf r(t)=10^u\\\sf\dfrac{dr}{du}=10^u\cdot\ell n(10)\\\sf \dfrac{du}{dt}=\diagup\!\!\!2\cdot\dfrac{1}{\diagup\!\!\!\!2\sqrt{t}}=\dfrac{1}{\sqrt{t}}$

$\sf\dfrac{dr}{dt}=\dfrac{dr}{du}\cdot\dfrac{du}{dt}\\\sf\dfrac{dr}{dt}=10^u\cdot\ell n(10)\cdot\dfrac{1}{\sqrt{t}}\\\tt mas~u=2\sqrt{t}~\rm ent\tilde ao:\\\sf\dfrac{dr}{dt}=10^{2\sqrt{t}}\cdot\ell n(10)\cdot\dfrac{1}{\sqrt{t}}\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf\dfrac{dr}{dt}=\dfrac{10^{2\sqrt{t}}\cdot\ell n(10)}{\sqrt{t}}\checkmark}}}}}$