Answer
$$h'(x)=\tan(2\sqrt x)+\sqrt x\sec^2(2\sqrt x)$$
Work Step by Step
$$h(x)=x\tan(2\sqrt x)+7$$
The derivative of $h(x)$ is: $$h'(x)=(x')\tan(2\sqrt x)+x\Big(\tan(2\sqrt x)\Big)'+0$$
$$h'(x)=\tan(2\sqrt x)+x\sec^2(2\sqrt x)(2\sqrt x)'$$
$$h'(x)=\tan(2\sqrt x)+x\sec^2(2\sqrt x)(2x^{1/2})$$
$$h'(x)=\tan(2\sqrt x)+x\sec^2(2\sqrt x)\Big(2\times\frac{1}{2}x^{-1/2}\Big)$$
$$h'(x)=\tan(2\sqrt x)+x\sec^2(2\sqrt x)\Big(\frac{1}{\sqrt x}\Big)$$
$$h'(x)=\tan(2\sqrt x)+\sqrt x\sec^2(2\sqrt x)$$
