Differentiation — practice questions

(a) Find the derivative $f'(x)$ We differentiate the function term by term using the power rule. $f(x) = 2x^3 - 5x^2 + 1x^1 + 9x^0$ $f'(x) = (3 \times 2)x^{3-1} - (2 \times 5)x^{2-1} + (1 \times 1)x^{1-1} + 0$ $f'(x) = 6x^2 - 10x + 1x^0$ Since $x^0 = 1$, we have: $f'(x) = 6x^2 - 10x + 1$

(b) Find the gradient at $x=2$ The gradient function $f'(x)$ tells us the gradient for any value of $x$. To find the gradient at $x=2$, we substitute $x=2$ into $f'(x)$. Gradient = $f'(2) = 6(2)^2 - 10(2) + 1$ $= 6(4) - 20 + 1$ $= 24 - 20 + 1$ $= 5$ So, the gradient of the curve at $x=2$ is 5.

The solution requires several steps:

1. Find the coordinates of the point. We are given $x=2$. Substitute this into the original equation to find $y$. $y = (2)^2 + \frac{8}{2} = 4 + 4 = 8$. The point of contact is P(2, 8).

2. Prepare the function for differentiation. Rewrite the equation using indices: $y = x^2 + 8x^{-1}$.

3. Differentiate to find the gradient function. $\frac{dy}{dx} = 2x + (-1) \times 8x^{-1-1}$ $\frac{dy}{dx} = 2x - 8x^{-2} = 2x - \frac{8}{x^2}$.

4. Find the gradient of the tangent at $x=2$. Substitute $x=2$ into $\frac{dy}{dx}$. $m_{\text{tangent}} = 2(2) - \frac{8}{2^2} = 4 - \frac{8}{4} = 4 - 2 = 2$.

5. Find the gradient of the normal. The normal is perpendicular to the tangent. $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{2}$.

6. Find the equation of the normal. Using the point P(2, 8) and $m = -\frac{1}{2}$ with the formula $y - y_1 = m(x - x_1)$. $y - 8 = -\frac{1}{2}(x - 2)$ To eliminate the fraction, multiply the entire equation by 2: $2(y - 8) = -1(x - 2)$ $2y - 16 = -x + 2$ Rearrange into the form $ax+by+c=0$. $x + 2y - 16 - 2 = 0$ $x + 2y - 18 = 0$