use part 1 of the fundamental theorem of calculus to find the derivative of the function. y = ∫₈^tan(x) √(3t…
use part 1 of the fundamental theorem of calculus to find the derivative of the function. y = ∫₈^tan(x) √(3t + √t) dt y =
Answer
Answer:
$\sqrt{3\tan(x)+\sqrt{\tan(x)}}\sec^{2}(x)$
Explanation:
Step1: Aplicar el Teorema Fundamental del Cálculo - Parte 1
Si $y = \int_{a}^{u(x)}f(t)dt$, entonces $y'=f(u(x))\cdot u'(x)$. Aquí, $a = 8$, $u(x)=\tan(x)$ y $f(t)=\sqrt{3t+\sqrt{t}}$.
Step2: Encontrar $u'(x)$
Como $u(x)=\tan(x)$, entonces $u'(x)=\sec^{2}(x)$.
Step3: Encontrar $f(u(x))$
Sustituir $t = u(x)=\tan(x)$ en $f(t)$. Tenemos $f(u(x))=\sqrt{3\tan(x)+\sqrt{\tan(x)}}$.
Step4: Calcular $y'$
Usando la fórmula $y'=f(u(x))\cdot u'(x)$, obtenemos $y'=\sqrt{3\tan(x)+\sqrt{\tan(x)}}\sec^{2}(x)$.