Differentiate function: $\ y=\sin x \sqrt{x+3}$
$\frac{1}{y}dy= \frac{1}{\sqrt{x+3}}dx + \frac{1}{\sin x}\cos xdx$
$\ln(y)=\ln(\sqrt{x+3})+\ln(\sin x)$
$\frac{dy}{dx}=y \left(\frac{1}{\sqrt{x+3}}+\frac{\cos x}{\sin x} \right)$
$\frac{dy}{dx}=\sqrt{x+3}\sin x \left(\frac{1}{\sqrt{x+3}}+\frac{\cos x}{\sin x} \right)$
$\frac{dy}{dx}=\sin x+\cos x\sqrt{x+3}$
But when I check with a product rule I get...
$\large \frac{dy}{dx}=\frac{\sin x}{2\sqrt{x+3}}+\cos x\sqrt{x+3}$
Where is a mistake?
1 Answer
I assume the function is $y=\sqrt{x+3}\sin x$. With the (formal) logarithmic derivative you have $$ \log y=\frac{1}{2}\log(x+3)+\log\sin x $$ so $$ \frac{y'}{y}=\frac{1}{2}\frac{1}{x+3}+\frac{\cos x}{\sin x} $$ and therefore $$ y'=\sqrt{x+3}\,\sin x\left(\frac{1}{2}\frac{1}{x+3}+\frac{\cos x}{\sin x}\right)= \frac{\sin x}{2\sqrt{x+3}}+\sqrt{x+3}\,\cos x $$ Note that $$ D(\log(\sqrt{x+3})=\frac{1}{\sqrt{x+3}}\frac{1}{2\sqrt{x+3}} $$ and you forgot the second factor (chain rule).
