Answer:
[tex]\dfrac{dy}{dx}=\dfrac{x}{4+x^{2}}[/tex]
Step-by-step explanation:
Given the function:
[tex]y= \ln{(4+x^2)^{\frac{1}{2}}[/tex]
as we know that [tex]\frac{d}{dx}(\ln{(f(x))})=\frac{1}{f(x)}(f'(x))[/tex]
[tex]y= \ln{(4+x^2)^{\frac{1}{2}}[/tex]
[tex]\dfrac{dy}{dx}= \dfrac{1}{(4+x^2)^{\frac{1}{2}}}\dfrac{d}{dx}((4+x^2)^{\frac{1}{2}})[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{1}{(4+x^2)^{\frac{1}{2}}}\left(\dfrac{1}{2}(4+x^2)^{\frac{1}{2}-1}(0+2x)\right)[/tex]
now we'll just simplify:
[tex]\dfrac{dy}{dx}=\dfrac{1}{(4+x^2)^{\frac{1}{2}}}\left(\dfrac{1}{2}(4+x^2)^{-\frac{1}{2}}(2x)\right)[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{x}{(4+x^{2})}[/tex]
this is our answer!
