1. **State the problem:** We need to find the gradient of the tangent to the curve $y=2x^3+5$ at the point $P(1,7)$ using differentiation from first principles. 2. **Recall the formula for differentiation from first principles:** $$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ This formula calculates the derivative as the limit of the average rate of change over an interval as the interval approaches zero. 3. **Apply the formula to our function:** Let $f(x) = 2x^3 + 5$. Calculate $f(1+h)$: $$f(1+h) = 2(1+h)^3 + 5 = 2(1 + 3h + 3h^2 + h^3) + 5 = 2 + 6h + 6h^2 + 2h^3 + 5 = 7 + 6h + 6h^2 + 2h^3$$ Calculate the difference quotient: $$\frac{f(1+h) - f(1)}{h} = \frac{(7 + 6h + 6h^2 + 2h^3) - 7}{h} = \frac{6h + 6h^2 + 2h^3}{h}$$ 4. **Simplify the difference quotient by canceling $h$:** $$\frac{\cancel{6h} + 6h^2 + 2h^3}{\cancel{h}} = 6 + 6h + 2h^2$$ 5. **Take the limit as $h$ approaches 0:** $$\lim_{h \to 0} (6 + 6h + 2h^2) = 6 + 0 + 0 = 6$$ 6. **Conclusion:** The gradient of the tangent to the curve at point $P(1,7)$ is $6$. **Final answer:** $$\frac{dy}{dx} = 6$$