Answer:
The derivative of the function is [tex]y'=\frac{1+\frac{2}{x}-ln5x}{x^2+4x+4}[/tex]
Step-by-step explanation:
Be
[tex]y=\frac{ln5x}{x+2}[/tex]
then, to derivate we have to apply the rule for division and the chain rule, and the division rule:
[tex]y'=\frac{d}{dx}(\frac{ln5x}{x+2})=\frac{\frac{d}{dx}(ln5x)(x+2)-(ln5x)\frac{d}{dx}(x+2)}{(x+2)^2}=\frac{(\frac{1}{x})(x+2)-ln5x(1)}{x^2+4x+4}=\frac{1+\frac{2}{x}-ln5x}{x^2+4x+4}[/tex]
which is our derivative, and because, by chain rule we calculated
[tex]\frac{d}{dx}(ln5x)=\frac{1}{5x}*5=\frac{1}{x}[/tex]
Answer:
[tex]y'=\frac{1+\frac{2}{x}-ln5x }{x^2+4x+4}[/tex]
Step-by-step explanation:
The function
[tex]y=\frac{ln5x}{x+2}[/tex]
its derivative can be found with the derivative rule of the quotient
if [tex]y=\frac{a}{b}[/tex]
the derivative is
[tex]y'=\frac{a'b-ab'}{b^2}[/tex]
in this case
[tex]a=ln5x[/tex]
and
[tex]b=x+2[/tex]
thus Using the logarithm derivation: [tex](lnu)'=\frac{u'}{u}[/tex], so
[tex]a'=\frac{5}{5x} =\frac{1}{x}[/tex]
and
[tex]b'=1[/tex]
the derivative of [tex]y=\frac{ln5x}{x+2}[/tex] is:
[tex]y'=\frac{a'b-ab'}{b^2}=\frac{(\frac{1}{x} )(x+2)-(ln5x)(1)}{(x+2)^2}\\y'=\frac{1+\frac{2}{x}-ln5x }{x^2+4x+4}[/tex]
