Derivative of a sum involving a root, power, and trig term

First rewrite $7\sqrt{x}$ as $7x^{1/2}$. Applying the power rule, $\frac{d}{dx}[7x^{1/2}] = 7\cdot\frac{1}{2}x^{-1/2} = \frac{7}{2}x^{-1/2}$. For the second term, $\frac{d}{dx}[-4x^2] = -4\cdot 2x = -8x$.

For the third term, recall $\frac{d}{dx}[\cos(x)] = -\sin(x)$, so $\frac{d}{dx}[3\cos(x)] = -3\sin(x)$. Adding these using the sum/difference rule gives $h'(x) = \frac{7}{2}x^{-1/2} - 8x - 3\sin(x)$, which can also be written as $\frac{7}{2\sqrt{x}} - 8x - 3\sin(x)$.

The term came from differentiating three cosine x, and since the derivative of cosine is negative sine, the constant multiple rule gives negative three sine x. So the sine term in the final answer is negative.