To find the derivate of a equation with natural logarithm you use the next:
[tex]\begin{gathered} y=\ln u \\ y^{\prime}=u^{\prime}\frac{1}{x} \end{gathered}[/tex]You need to turn the given equation as follow:
[tex]\begin{gathered} y=\ln (x^2+3) \\ y=\ln u \\ u=x^2+3 \end{gathered}[/tex]Find the derivate of u:
[tex]\begin{gathered} y=x^n \\ y^{\prime}=nx^{n-1} \end{gathered}[/tex][tex]\begin{gathered} u^{\prime}=2x+0 \\ u`=2x \end{gathered}[/tex]Then, you find the derivate of the natural logarithm:
[tex]\begin{gathered} y=\ln (x^2+3) \\ y^{\prime}=u^{\prime}\cdot\frac{1}{x^2+3} \\ \\ y^{\prime}=\frac{2x}{x^2+3} \end{gathered}[/tex]The answer is:
[tex]y^{\prime}=\frac{2x}{x^2+3}[/tex]-------------------------------------------------
[tex]y=2x\sin x[/tex]You use the next formula for the derivate of a product:
[tex]\begin{gathered} y=xg \\ y^{\prime}=xg^{\prime}+gx^{\prime} \end{gathered}[/tex][tex]\begin{gathered} y=\sin x \\ y^{\prime}=\cos x \end{gathered}[/tex]You get:
[tex]\begin{gathered} y^{\prime}=2x\cos x+\sin x(2x^0) \\ y^{\prime}=2x\cos x+2\sin x \end{gathered}[/tex]In x=π/2
[tex]\begin{gathered} y^{\prime}=2(\frac{\pi}{2})\cos (\frac{\pi}{2})+2\sin (\frac{\pi}{2}) \\ \\ y^{\prime}=\pi\cos \frac{\pi}{2}+2\sin \frac{\pi}{2} \\ y^{\prime}=\pi(0)+2(1) \\ \\ y^{\prime}=2 \end{gathered}[/tex]The answer is
[tex]y^{\prime}=2[/tex]