Question

Use a regra da cadeia e calcule a derivada das funções:
a) f(x) = Ln(x^2 - 2x)
c) f(x) = 5£^4-x^2
d) f(x) = x^2£^x^2​

Answer 1

a)

$\frac{d\left(ln\left(x^2-2x\right)\right)}{dx}~=\\\\\\Sendo~~u(x)=x^2-2x\\\\\\=~\frac{d\left(ln\left(u(x)\right)\right)}{dx}\\\\\\=~\frac{d\left(ln(u(x)\right)}{du}\,.\,\frac{d(u(x))}{dx}\\\\\\=~\frac{1}{u(x)}~.~(2x-2)\\\\\\=~\frac{1}{x^2-2x}~.~(2x-2)\\\\\\=~\boxed{\frac{2x-2}{x^2-2x}}$

As letra (b) e (c) estão confusas, vou resolve-las como acredito ser o certo, já que não é possível identificar com certeza o que é expoente nas expressões.

b)

$\frac{d\left(5e^{4-x^2}\right)}{dx}~=\\\\\\Sendo~~u(x)=4-x^2\\\\\\=~\frac{d\left(5e^{u(x)}\right)}{dx}\\\\\\=~\frac{5e^{u(x)}}{du}~.~\frac{du(x)}{dx}\\\\\\=~5e^{u(x)}~.~(-2x)\\\\\\=~5e^{4-x^2}~.~-2x\\\\\\=~\boxed{-10x.e^{4-x^2}}$

c)

$\frac{d\left(x^2e^{x^2}\right)}{dx}~=\\\\\\Utilizando~primeiro~a~regra~do~produto\\\\\\=~\frac{d\left(x^2\right)}{dx}~.~e^{x^2}~+~\frac{d\left(e^{x^2}\right)}{dx}~.~x^2\\\\\\=~2x.e^{x^2}~+~x^2\,.\,\frac{d\left(e^{x^2}\right)}{dx}$

Vamos calcular separadamente a derivada restante pela regra da cadeia

$\frac{d\left(e^{x^2}\right)}{dx}~=\\\\\\Sendo~~u(x)=x^2\\\\\\=~\frac{d\left(e^{u(x)}\right)}{dx}\\\\\\=~\frac{d\left(e^{u(x)}\right)}{du}~.~\frac{du(x)}{dx}\\\\\\=~e^{u(x)}~.~2x\\\\\\=~\boxed{2x.e^{x^2}}$

Substituindo no resultado obtido pela regra do produto:

$=~2x.e^{x^2}~+~x^2\,.\,2x\,.\,e^{x^2}\\\\\\=~2x.e^{x^2}~+~2x^3.e^{x^2}\\\\\\=~\boxed{(1+x^2)\,.\,2x.e^{x^2}}$