¡GoConqr estrena diseño! Cuéntanos qué te parece.

Apuntes

MEI OCR: Core 3 Maths

Sophie.J.E Maths A-Levels Inglés 1 página

Contenido del apunte (1)

Página 1

Table of Differential functions

y dy/dx

x^n nx^n-1

ax^n anx^n-1

sin(ax) acos(ax)

cos(ax) -asin(ax)

tan(x) 1/cos^2(x) (a.k.a sec^2(x))

e^ax ae^ax

alnx a/x

Chain rule - Used for functions of functions

dy/dx=(du/dx)(dy/du)

E.g. Find dy/dx when x=1 for y=1/sqrt(x^2+4x):

u=x^2+4x, y=u^-(1/2), du/dx=2x+4, dy/du=-(1/2)u^-(3/2)

dy/dx=-(1/2)(x^2+4x)^-(3/2) (2x+4) = -(x+2)/sqrt(x^2+4x)^3

Product rule - Used to differentiate one function divided by another

dy/dx= v(du/dx)+u(dv/dx)

E.g. Differentiate x^3tanx with respect to x:

u=x^3, v=tanx, du/dx=3x^2, dv/dx=1/(cos^2(x))

dy/dx=x^3/(cos^2(x))+3x^2(tanx)

Differentiation Rules

Quotient rule - Used to differentiate two finctions which are multiplied together

dy/dx=(v(du/dx)-u(dv/dx))/v^2

E.g. Differentiate y=cos(x)/sin(x) with respect to x:

u=cos(x), v=sin(x), du/dx=-sin(x), dv/dx=cos(x)

dy/dx=(-sin^2(x)-cos^2(x))/sin^2(x) = -1/sin^2(x)

Implicit Differentiation - If you can't write the equation as y=f(x), then use implicit differentiation. To do this, you first differentiate terms of x with respect to x, as you normally would. Then, use the chain rule to differentiate terms of y only (in other words, differentiate the y terms with respect to y and place a dy/dx after them). Use the product rule to differentiate in terms of x and y if need be and rearrange the final equation to make dy/dx the subject.

E.g. Differentiate 2x^2-4y^3=6

4x-12y^2(dy/dx)=0

3y^2(dy/dx)=x

dy/dx=x/3y^2

Related Rates of Change

Some situations have a number of linked variables, like length, surface area and volume, or distance speed and acceleration. If you know the rate of change of one of these variables, and the equations that connect the variables, you can use the chain rule to help you find the rate of change of one of the other variables.

E.g. A giant metal cube from space is cooling after entering the Earth's atmosphere. As it cools, the surface area of the cube decreases at a constant rate of 0.027m^2 s^-1. If the side length of the cube after t seconds is xm, find dx/dt at the point when x=15m.

The cube has a side length of xm, so the surface area of the cube is A=6x^2 therefore dA/dx=12x. A decreases at a constant rate of 0.027m^2 s^-1 so dA/dt=-0.027 (its a minus because it is decreasing)

We need to find dx/dt so we use the chain rule: dx/dt=(dx/dA)(dA/dt)

dx/dt=1/(dA/dx) (dA/dt)=-0.027/12x=-0.00225/x

So when x=15, dx/dt=-0.00225/15=-0.00015m s^-1

Table of Integrals

f(x) Integral of f(x)

x^n (1/n+1)x^(n+1)

ax^n (a/n+1)x^(n+1)

sinax -(1/a)cosax

cosax (1/a)sinax

a/x alnx

e^x

Differentiation

Integration

Información

Descripción
A-Levels Maths Apunte sobre MEI OCR: Core 3 Maths, creado por Sophie.J.E el 25/10/2013.
Creado por
Sophie.J.E · oct 2013
Actualizado
23 nov 2016
Visitas
322
Etiquetas
maths
Valoración

Escríbenos

¿Sobre qué es tu mensaje?

¿Qué te parece el nuevo diseño?

¿Te gusta?