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KathillyRayanne
@KathillyRayanne
December 2019
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Qual a derivada da função :
f( x ) = ( x + 1 )^2 / 1 + x^2
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Ruthyafiel
Se f(x)=1/x² => f(x)= x^(-2)
Pela regra geral , f(x)= a^n => f'(x)=n.a^(n-1),
temos:
df/dx= -2x^(-3),
que ^ indica elevado
a.. tpw, x^2=x².
f(x)=1/x²
f'(x)=[1'x²-1.x²']/(x²)²
f'(x)=[0.x²-1.2x]/x^4
f'(x)=-2x/x^4
f'(x)=-2/x³
0 votes
Thanks 1
bruno030307
a sua faroração da função estar errada f(x) nao é 1 / x^2
Ruthyafiel
sorry my glue was wrong rsrsrs in short summing up of 10 am 4 in number
bruno030307
0 votes
Thanks 1
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Lista de comentários
Pela regra geral , f(x)= a^n => f'(x)=n.a^(n-1),
temos:
df/dx= -2x^(-3),
que ^ indica elevado
a.. tpw, x^2=x².
f(x)=1/x²
f'(x)=[1'x²-1.x²']/(x²)²
f'(x)=[0.x²-1.2x]/x^4
f'(x)=-2x/x^4
f'(x)=-2/x³