Answer:
[tex]\huge\boxed{f'(\sqrt{2x})=-2\sqrt{2x}\cos2x}[/tex]
Step-by-step explanation:
[tex]f(x)=\sin(x^2+\pi)\\\\f'(x)=\bigg(\sin(x^2+\pi)\bigg)'\\\\\text{use}\\(\sin x)'=\cos x\\\bigg(f\left(g(x)\right)\bigg)'=f'\left(g(x)\right)\cdot g'(x)\\\\f'(x)=\cos(x^2+\pi)\cdot(x^2+\pi)'\\\\\text{use}\\\bigg(x^n\bigg)'=nx^{n-1}\\(c)'=0\\\\f'(x)=\cos(x^2+\pi)\cdot2x=2x\cos(x^2+\pi)[/tex]
[tex]f'(\sqrt{2x})-\text{substitute}\ x=\sqrt{2x}\ \text{into}\ f'(x):\\\\f'(\sqrt{2x})=2(\sqrt{2x})\cos\bigg((\sqrt{2x})^2+\pi\bigg)=2\sqrt{2x}\cos(2x+\pi)=-2\sqrt{2x}\cos2x\\\\\text{used}\ \cos(\alpha+\pi)=-\cos\alpha[/tex]
