## The Chain Rule

The chain rule allows you to differentiate composite functions (functions of other functions) ie) f(g(x)) such as sin(3x^{2}) or (5x^{3}+2x+3)^{2}. The rule is as follows

or to understand it more simply you differentiate the inner function and multiply it by the derivative of the outer function (leaving what’s inside alone).

## Differentiating brackets raised to a power

The chain rule can be a great short cut to differentiating brackets raised to a power as it doesn’t require you to multiply them all out, it also enables you to differentiate brackets raised to an unknown power.

Consider

This is the composite of the functions ax+b and t^{n}. So we differentiate them both to get a and nt^{n-1} and then apply the formula to get

Notice how we multiplied the derivative of the inner function, a, by the derivative of the outer function nt^{n-1} but substituted ax+b back in for t.

To generalise we can replace the ax+b with f(x) and by applying the above get

## Differentiating Trigonometric functions

We can also use the chain rule when differentiating sin(f(x)) and cos(f(x)) since we know how to differentiate sin(x) and cos(x).

Using the chain rule we get

and

## Compound Angles – sin(A+B) = cosAsinB+sinAcosB

Compound angles are angles made by adding two other angles together. When using trigonometry unfortunately you cant just “times out” the trig function but have to use an identity. This post will consider how we get the identity for sin(A+B):

**sin(A+B) = sinAcosB+sinBcosA**

From the definition of sin=opp/hyp we find

sin(A+B) = RT/OR

But sinceRT comprises of RS+ST

By David Woodford

## Sec, Cosec, Cot

Sec, cosec and cot are all functions in trigonometry. They are simply equal to one over on of the other functions, ie cos, sin and tan.

so

Sec = 1/cos

Cosec = 1/sin

cot = 1/tan

You can remember which is paired with which using the third letter rule. This is that the third letter is the first letter of the corresponding function ie)

se**c **goes with **c**os

co**s**ec goes with **s**in

co**t** goes with **t**an

## Differentiate Inverse Cos – Proof

**Now available from trevorpythag.blogspot.com**

How to differentiate cos^{-1}x

y=cos^{-1}x

Bring the cos across

cosy = x

Differentiate both sides, remember when differentiating y time by dy/dx

-sin(y) dy/dx = 1

dy/dx = -1/siny

However we want to get the differential in terms of x, to do this we can use the identity

sin^{2}t+cos^{2}t = 1

so

sint = √(1 – cos^{2}t)

putting this into our expression for dy/dx we get

dy/dx = -1/√(1-cos^{2}y)

but cosy = x so

**dy/dx =- 1/√(1-x ^{2})**

## Trigonometry Identities

There a number of “identities” in trigonometry that can be found from the basic ideas of sin, cos and tan as explained in my earlier post. These identities can help in solving equations involving trig functions, especially when there are 2 or more different functions as the often allow you to write the equation in terms of one function, eg sin, that you can then solve. One of the identities is: **sin ^{2} + cos^{2} = 1. ** To prove this consider a right angled triangle with side a,b and c as shown below

From this we can use Pythagoras theorem to say: a^{2}+b^{2}=c^{2 now we know sin t = b/c so b = csin t cos t = a/c so a = ccos t} substituting these values in the above equation we get (csint)^{2} +(ccost)^{2} = c^{2} canceling the c^{2} we get **sint ^{2} + cost^{2} = 1**

There are trig functions that are equal to 1 over sin, cos and tan called cosec = 1/sin, sec = 1/cos and cot = 1/tan. These can be remembered using the third letter rule as the third letter of each of these corresponds to the the function it is one over.

Using these a cos

^{2}+ sin

^{2}= 1 we can calculate other identities

**tan**We can obtain this by dividing through by cos

^{2}t + 1 = sec^{2}t^{2}as we know sin/cos = tan, cos/cos = 1 and 1/cos = sec. Other similar identities can be obtained for cosec and cot.

## Tan = sin/cos

this site is now at www.breakingwave.co.nr

This is often useful when solving trig equations so i thought i’d include it

basically:

sin = opp/hyp

and

cos=adj/hyp

so

sin/cos = ^{(opp/hyp)}/_{(adj/hyp)}

so if we cancel the hyp’s we get

sin/cos = opp/adj

and since tan = opp/adj

**tan = sin/cos**

## Sine and Cos Graphs Differentiating sin and cos

This is the basics of the sine cos and tan graphs and how sine and cos relate to give you tan. It also shows how to differentiate sin and cos.

The output or range of both sine and cos is from -1 to 1 when given any angle. They can be shown on a graph where y = sin(x) and y = cos(x). In these graphs all the angles go along the x axis and you can see a wave type shape is formed

**Sine Graph**

**Cosine Graph**

As you can see both the sin and cos graphs move periodically between -1 and 1 as the angles change, this pattern continues indefinitely because once you pass 360 degrees or 2 pi radians you will return back to the beginning. If you try to perform sin^{-1} of a value out side the range -1 to 1 you will get an error.

**Differentiate Sin and Cos**

also notice that the gradient of the sin graph is the value of the cos graph for the same angle and that the gradient of the cos graph is the -value of the sin graph for that angle. This means that we can differentiate the sin and cos graphs:

if f(x) = sin(x) then f ‘ (x)=cos(x)

and

if f(x) = cos(x) then f ‘ (x) = -sin(x)

however if we use ax instead of x we must differentiate it by bringing the a out, when its just x this doesn’t matter as the differential of x is 1.

ie)

let y = sin(f(x))

now let u = f(x)

du/dx = f ‘ (x)

also

y=sin(u) as u = f(x)

dy/du = cos(u)

from the chain rule

dy/dx = du/dx * dy/du

therefore

**if y = sin(f(x))
dy/dx = f ‘ (x)cos(f(x))**

**and similarly for cos
if y = cos(f(x))
dy/dx = -f ‘ (x)sin(f(x))**

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