Question

d) $\frac{d}{dx}(6 + 7t+8t^{2}) = 0$

Explanation:

Step1: Apply sum - rule of differentiation

The sum - rule states that $\frac{d}{dx}(u + v+w)=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}$. Here $u = 6$, $v = 7t$, $w = 8t^{2}$. So $\frac{d}{dx}(6 + 7t+8t^{2})=\frac{d}{dx}(6)+\frac{d}{dx}(7t)+\frac{d}{dx}(8t^{2})$.

Step2: Differentiate each term

Since $6$, $7t$, and $8t^{2}$ are treated as constants with respect to $x$ (because there is no $x$ in them), the derivative of a constant $C$ with respect to $x$ is $0$. So $\frac{d}{dx}(6)=0$, $\frac{d}{dx}(7t)=0$, $\frac{d}{dx}(8t^{2}) = 0$. Then $\frac{d}{dx}(6)+\frac{d}{dx}(7t)+\frac{d}{dx}(8t^{2})=0 + 0+0=0$.

Answer:

The left - hand side $\frac{d}{dx}(6 + 7t+8t^{2}) = 0$ holds true because $6+7t + 8t^{2}$ is a constant with respect to $x$, and the derivative of a constant with respect to a variable is $0$.