Algebra and Graphs - Basic differentiation and gradients

Differentiate $y = 5x^3 - 2x^2 + 4x - 7$ with respect to $x$.

$\frac{dy}{dx} = 15x^2 - 4x + 4$

Apply the power rule to each term: $3 \times 5x^{(3-1)} = 15x^2$, $2 \times 2x^{(2-1)} = 4x$, the derivative of $4x$ is $4$, and the derivative of the constant $-7$ is $0$.

Find the gradient of the curve $y = x^2 + 3x$ at the point where $x = 2$.

$7$

First, find the derivative: $\frac{dy}{dx} = 2x + 3$. To find the gradient at the specific point $x = 2$, substitute $2$ into the derivative: $2(2) + 3 = 4 + 3 = 7$.

Find the coordinates of the stationary point on the curve $y = x^2 - 6x + 5$.

$(3, -4)$

1. Find $\frac{dy}{dx} = 2x - 6$. 2. Set $\frac{dy}{dx} = 0$ for stationary points: $2x - 6 = 0 \implies x = 3$. 3. Substitute $x = 3$ back into the original equation to find $y$: $y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4$.

1. Find the $y$-coordinate when $x = 3$: $y = (3)^2 - 4 = 9 - 4 = 5$ Point is $(3, 5)$.

2. Find the derivative $\frac{dy}{dx}$: $\frac{dy}{dx} = 2x$

3. Calculate gradient $m$ at $x = 3$: $m = 2(3) = 6$

4. Use $y - y_1 = m(x - x_1)$: $y - 5 = 6(x - 3)$ $y - 5 = 6x - 18$ $y = 6x - 13$

First find the point on the curve, then find the gradient function using differentiation. Substitute the $x$-value to find the specific gradient and use the point-slope formula for the line.

1. Find $\frac{dy}{dx}$: $\frac{dy}{dx} = 12 - 6x$

2. Set $\frac{dy}{dx} = 0$ for stationary points: $12 - 6x = 0 \implies 6x = 12 \implies x = 2$

3. Find $y$-coordinate: $y = 12(2) - 3(2)^2 = 24 - 12 = 12$ Point is $(2, 12)$.

4. Determine nature using $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = -6$ Since $-6 < 0$, the point $(2, 12)$ is a Maximum.

Differentiate to find the gradient function. Solve for $x$ when the gradient is zero. Check the sign of the second derivative to classify the point as a maximum or minimum.