Calcular derivada da função dada
12 set 2017, 05:19
Calcule a derivada da função:
\(y=\frac{2x-3}{5x+4}\)
\(y=\frac{2x-3}{5x+4}\)
Re: Calcular derivada da função dada
12 set 2017, 16:07
\(\frac{d}{dx}(\frac{u}{v})=\frac{vu'-uv'}{v^2}\)
\(u=2x-3\rightarrow u'=2\\ v=5x+4\rightarrow v'=5\\ v^2=(5x+4)^2\)
\(\frac{vu'-uv'}{v^2}=\frac{(5x+4)2-(2x-3)5}{(5x+4)^2}=\frac{10x+8-(10x-15)}{(5x+4)^2}=\frac{10x+8-10x+15}{(5x+4)^2}=\frac{13}{(5x+4)^2}\)
\(u=2x-3\rightarrow u'=2\\ v=5x+4\rightarrow v'=5\\ v^2=(5x+4)^2\)
\(\frac{vu'-uv'}{v^2}=\frac{(5x+4)2-(2x-3)5}{(5x+4)^2}=\frac{10x+8-(10x-15)}{(5x+4)^2}=\frac{10x+8-10x+15}{(5x+4)^2}=\frac{13}{(5x+4)^2}\)