[list][*]Understand the gradient of a curve at a point as the limit of the gradients of a suitable sequence of chords, and use the notations for first and second derivatives.[/*][*]Use the derivative of x^n (for any rational n), together with constant multiples, sums and differences of functions, and of composite functions using the chain rule.[/*][*]Apply differentiation to gradients, tangents and normals.[/*][/list]

The basic concept of differentiation is denoted below:[br][table][tr][td][math]y[/math][/td][td][math]\frac{dy}{dx}[/math][/td][/tr][tr][td][math]a\cdot x^n[/math][/td][td][math]a\cdot n\cdot x^{n-1}[/math][/td][/tr][tr][td][math]4x^2[/math][/td][td][math]8x[/math][/td][/tr][tr][td][math]1000x[/math][/td][td][math]1000[/math][/td][/tr][tr][td][math]10^{10}[/math][/td][td][math]0[/math][/td][/tr][/table][br]Chain Rule:[br]If [math]y=u[/math]; [math]u=f\left(x\right)[/math], then [math]\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}[/math][br]Another representation is: [math]y=a\left[f\left(x\right)\right]^n[/math], so [math]\frac{dy}{dx}=a\cdot n\cdot\left[f\left(x\right)\right]^{n-1}\cdot f'\left(x\right)[/math][br]E.g. Find [math]\frac{dy}{dx}[/math] of [math]y=\left(x^3+2x-1\right)^2[/math][br]Solution 1: [br]Let [math]y=u^2[/math] and [math]u=x^3+2x-1[/math][br]Now, [math]\frac{dy}{du}=2u[/math] and [math]\frac{du}{dx}=3x^2+2[/math][br]So that, [math]\frac{dy}{dx}=2\left(x^3+2x-1\right)\cdot\left(3x^2+2\right)[/math][br]Solution 2:[br][math]\frac{dy}{dx}=2\left(x^3+2x-1\right)^{2-1}\cdot\left(3x^2+2\right)[/math][br][br]As for a brief explanation, we can just know that gradient is defined as [math]m=\frac{y_2-y_1}{x_2-x_1}[/math].[br]From there, we can take that [math]m=\frac{dy}{dx}[/math]. Thus, we can make equations of tangent and normal from a curve.[br]Equation of tangent: [math]y=mx+c[/math] or [math]y-y_1=m\left(x-x_1\right)[/math][br]Equation of normal: [math]y=-\frac{1}{m}x+c[/math] or [math]y-y_1=-\frac{1}{m}\left(x-x_1\right)[/math][br]Another thing to add is whenever the, [math]m=\frac{dy}{dx}=0[/math][math][/math], we can look for stationary points of the curve.