Question

Find the slope of the tangent to the curve f (x) = 2x⁶ + x⁴ − 1 at x = 1.

Answer 1

$$\begin{gathered} \text{Slope of the tangent =}f\text{'}\left(x\right) \\ =\frac{d}{dx}\left(2x^6+x^4-1\right) \\ =2\frac{d}{dx}\left(x^6\right)+\frac{d}{dx}\left(x^4\right)-\frac{d}{dx}\left(1\right) \\ =12x^5+4x^3 \\ \therefore \text{Slope of the tangent at}\mathit{\text{x}}\text{=1:} \\ 12{\left(1\right)}^5+4{\left(1\right)}^3=12+4=16 \end{gathered}$$