I've come across this derivation that looks fairly simple yet I cannot get it right.We are given solutions but I dont understand it.The function looks like this
$$f(x) = \cos y \space \ln x + y\sqrt{x^2 +3} $$
Now since $\cos y$ and $\ln x$ are multiplied and $y$ and the square term as well than we use the multiplication rule. The given solution looks like this
$$ -\sin y*y' \ln x + \cos y \frac 1x + y'\sqrt{x^2 +3} + y\space \frac 1 2 \space \frac {1}{x^2 +3}\space 2x$$
The part with the first multiplication makes sence, we derivate sin and ln x, we leave y as it is since x is the differation variable but how does one get to $$ \frac 1 2 \space \frac {1} {x^2 +3}\space 2x $$ from the root? If we would to get ride of the root by making it $$(x^2 + 3)^{1/2} $$ than we would have to use the power rule, which is bring 1/2 infront than we have to substract 1 from 1/2 so that is -1/2. But how does the 2x appear? Im aware that the $x^2 + 3$ under the root were also derivated but why?I thought the chain rule only applied to multiplication.A litle insight would be great,thanks!
There's a typo in the book.
$$\frac{d}{dx}(y\sqrt{x^2+3}) = y'\sqrt{x^2+3} + \frac{y}{2} \frac{1}{\color{blue}{\sqrt{x^2+3}}}2x$$
Now for the derivative of $\sqrt{x^2+3}$
$$\frac{d}{dx}\sqrt{x^2+3} = \left[\frac{d}{d(\color{red}{x^2})} \sqrt{\color{red}{x^2}+3}\right]\times\left[\frac{d(\color{green}{x^2})}{d\color{green}x}\right] = \frac{1}{2\sqrt{x^2+3}}\times (2x)$$