• Revision:Chain, Product and Quotient

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The Chain Rule

The chain rule is very important in differential calculus and states that:

\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}

This rule allows us to differentiate a vast range of functions.

Proof

Suppose that y is a function of t, and t is a function of x.

If \delta y, \delta t and \delta x are corresponding increments in the variables in the variables y, t andx, then

\frac{\delta y}{\delta x}=\frac{\delta y}{\delta t}\times \frac{\delta t}{\delta x} (1)

When \delta y, \delta t and \delta x tend to zero,

\frac{\delta y}{\delta x} \rightarrow \frac{dy}{dx}

\frac{\delta y}{\delta t} \rightarrow \frac{dy}{dt}

\frac{\delta t}{\delta x} \rightarrow \frac{dt}{dx}

so equation (1) becomes

\frac{dy}{dx}=\frac{dy}{dt} \times \frac{dt}{dx}

Examples

If y=(1+x^2)^3, find \frac{dy}{dx}.

Let t=1+x^2

\frac{dt}{dx}=2x

y=t^3

\frac{dy}{dt}=3t^2

By the Chain Rule, \frac{dy}{dx}=\frac{dy}{dt} \times \frac{dt}{dx}.

\Rightarrow \frac{dy}{dx}=3t^2 \times 2x= 3(1 + x^2)^2 \times 2x

\frac{dy}{dx}=6x(1 + x^2)^2

In examples such as the above one, with practice it should be possible for you to be able to simply write down the answer without having to let t=1 + x^2 etc. This is because:

When y=(f(x))^n

\frac{dy}{dx}=f'(x)\times n \times (f(x))^{(n-1)}

In other words, the differential of something in a bracket raised to the power of n is the differential of the bracket, multiplied by n multiplied by the contents of the bracket raised to the power of (n-1).


The Product Rule

This is another very useful formula, when we have two functions u and v, multiplied together:

\frac{d}{dx}(uv)=v\frac{du}{dx}+u\frac{dv}{dx}

Proof

Consider y=f(x) \times g(x).

Then y=uv (1)

Let x increase by a small amount \delta x, with corresponding increases in y, u and v of \delta y, \delta u and \delta v, so that

y+\delta y=(u+\delta u)(v+\delta v)

\Rightarrow y + \delta y=uv+u\delta v+v\delta u+\delta u\delta v (2)

Then, subtracting (1) from (2).

\delta y = u\delta v + v\delta u+\delta u\delta v

Then, dividing by \delta x,

\frac{\delta y}{\delta x}=\frac{u\delta v}{\delta x}+ \frac{v\delta u}{\delta x}+\frac{\delta u\delta v}{\delta x}

\Rightarrow \frac{\delta y}{\delta x}=u\frac{\delta v}{\delta x}+v\frac{\delta u}{\delta x}+\delta u\frac{\delta v}{\delta x}

Let \delta x \rightarrow 0.

Then \delta u \rightarrow 0 and \delta v \rightarrow 0,

\frac{\delta y}{\delta x} \rightarrow \frac{dy}{dx}

\frac{\delta v}{\delta x} \rightarrow \frac{dv}{dx}

\frac{\delta u}{\delta x} \rightarrow \frac{du}{dx}

\Rightarrow \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}+0\times\frac{dv}{dx}

\Rightarrow \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}

Examples

Differentiate x(x^2+1)

Let u=x and v=x^2+1

\Rightarrow \frac{du}{dx}=1 and \frac{dv}{dx}=2x

\frac{d}{dx}(x(x^2+1))=(x^2+1)\times 1 + x\times 2x

\frac{d}{dx}(x(x^2+1))=x^2+1+2x²

\Rightarrow \frac{d}{dx}(x(x^2+1))=3x^2+1

Again, with practice you shouldn't have to write out u = ... and v = ... every time.

The Quotient Rule

This formula lets us differentiate two functions divided by each other.

\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

Proof

Writing u=f(x) and v=g(x), then

y=\frac{u}{v} (1)

and

y+\delta y=\frac{u+\delta u}{v+\delta v} (2)

Substituting (1) into (2),

\delta y=\frac{u+\delta u}{v+\delta v}-\frac{u}{v}

\delta y=\frac{v(u+\delta u)-u(v+\delta v)}{v(v+\delta v)}

\delta y=\frac{vu+v\delta u-uv-u\delta v}{v^2+v\delta v}

Dividing by \delta x,

\frac{\delta y}{\delta x}=\frac{v\frac{\delta u}{\delta x}-u\frac{\delta v}{\delta x}}{v^2+v\delta v}

Let \delta x \rightarrow 0, then \delta u \rightarrow 0, \delta v \rightarrow 0 and

\frac{\delta y}{\delta x}\rightarrow \frac{dy}{dx}

\frac{\delta v}{\delta x}\rightarrow \frac{dv}{dx}

\frac{\delta u}{\delta x}\rightarrow \frac{du}{dx}

\Rightarrow \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

Examples

If y=\frac{x^3}{x+4}, find \frac{dy}{dx}.

Let u=x^3 and v=x+4.

\Rightarrow \frac{du}{dx}=3x^2 and \frac{dv}{dx}=1

\frac{d}{dx}(\frac{x^3}{x+4})=\frac{(x+4)\times (3x^2)-(x^3)\times (1)}{(x+4)^2}

\Rightarrow \frac{d}{dx}(\frac{x^3}{x+4})=\frac{2x^3+12x^2}{(x+4)^2}


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