Question

Preciso de ajuda para responder, se possível com resolução, por favor!

Answer 1

Boa tarde!

Quantas derivadas!
a)
$f(x)=x^2+x\\ f'(x)=2x+1\\ f'(1)=2(1)+1=2+1=3$

b)
$f(x)=\sqrt{x}\\ f(x)=x^{1/2}\\ f'(x)=(1/2)x^{1/2-1}\\ f'(x)=(1/2)x^{-1/2}\\ f'(x)=\frac{1}{2\sqrt{x}}\\ f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}$

c)
$f(x)=5x-3\\ f'(x)=5\\ f'(-3)=5$

d)
$f(x)=\frac{1}{x}\\ f(x)=x^{-1}\\ f'(x)=(-1)x^{-1-1}\\ f'(x)=\frac{-1}{x^2}\\ f'(1)=\frac{-1}{1^2}=-1$

e)
$f(x)=\sqrt{x}\\ f(x)=x^{1/2}\\ f'(x)=(1/2)x^{1/2-1}\\ f'(x)=(1/2)x^{-1/2}\\ f'(x)=\frac{1}{2\sqrt{x}}\\ f'(3)=\frac{1}{2\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{6}$

f)
$f(x)=\frac{1}{x^2}\\ f(x)=x^{-2}\\ f'(x)=(-2)x^{-2-1}\\ f'(x)=\frac{-2}{x^3}\\ f'(2)=\frac{-2}{2^3}=\frac{-1}{4}$

g)
$f(x)=2x^3-x^2\\ f'(x)=6x^2-2x\\ f'(1)=6(1)^2-2(1)=6-2=4$

h)
$f(x)=\sqrt[3]{x}\\ f(x)=x^{1/3}\\ f'(x)=(1/3)x^{1/3-1}\\ f'(x)=(1/3)x^(-2/3)\\ f'(x)=\frac{1}{3\sqrt[3]{x^2}}\\ f'(2)=\frac{1}{3\sqrt[3]{2^2}}=\frac{1}{3\sqrt[3]{4}}$

Espero ter ajudado!