Question

derivada de
a)$y=3x$
b)$y=(2)^-^x$
c)$y =e^2^x$
d)$y=e^-^3^x$
e)$y=e^l^n^x$

Answer 1

a)

y = 3x

y' = 3

----------------------

b)

$\\ y' = 2^-^x \\ \\ y' = 2^-^x*-1*ln2 \\ \\ y' = -2^-^xln2$

Obs: derivada de 2^x = 2^(x)*ln2, mas como a expoente -1, usa a regra da cadeia.

-------------------------------------

$\\ c) \\ \\ y = e^2^x \\ \\ y' = e^2^x*(2x)' \\ \\ y' = 2e^2^x$


derivada de e^x é ele mesmo. Só mudou porque tinha que usar a regra da cadeia.

-----------------------------------------


$\\ d) \\ \\ y = e^-^3^x \\ \\ y' = (e^-^3^x)*(-3x)' \\ \\ y' = e^-^3^x*(-3) \\ \\ y' = -3*e^-^3^x$

-----------------------------------------


$\\ e) \\ \\ y = e^l^n^(^x^) \\ \\ y' = e^l^n^(^x^)*(lnx)' \\ \\ y' = e^l^n^(^x^)* \frac{1}{x} \\ \\ y' = \frac{e^l^n^(^x^)}{x}$

Answer 2

$\sf \bullet A \to$

$\sf \displaystyle y=3x\\\\\\\frac{d}{dx}\left(3x\right)\\\\\\=3\frac{d}{dx}\left(x\right)\\\\\\=3\cdot \:1\\\\\\\to \boxed{\sf =3}$

$\sf \bullet B\to \\\\\\\displaystyle y=\left(2\right)^{-x}\\\\\\=\frac{d}{dx}\left(e^{\left(-x\right)\ln \left(2\right)}\right)\\\\\\=e^{\left(-x\right)\ln \left(2\right)}\frac{d}{dx}\left(\left(-x\right)\ln \left(2\right)\right)\\\\\\=e^{\left(-x\right)\ln \left(2\right)}\left(-\ln \left(2\right)\right)\\\\\\\to \boxed{\sf =-\ln \left(2\right)\cdot \:2^{-x}}$

$\sf \bullet C \to \\\\y=e^{2x}\\\\\\\dfrac{d}{dx}\left(e^{2x}\right)\\\\\\=e^{2x}\dfrac{d}{dx}\left(2x\right)\\\\\\\to \boxed{\sf =e^{2x}\cdot \:2}$

$\sf \bullet D\to \\\\\\y=e^{-3x}\\\\\\\dfrac{d}{dx}\left(e^{-3x}\right)\\\\\\=e^{-3x}\dfrac{d}{dx}\left(-3x\right)\\\\\\=e^{-3x}\left(-3\right)\\\\\\\to \boxed{\sf =-3e^{-3x}}$

$\sf \bullet E \to\\\\\\\:y=e^{inx}\\\\\\\displaystyle \frac{d}{dx}\left(e^{inx}\right)\\\\\\=e^{inx}\frac{d}{dx}\left(inx\right)\\\\\\\to \boxed{\sf =e^{inx}in}$