The Chain Rule | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Account for both changes

In $y=(3x-2)^7$ , first calculate $3x-2$ , then raise the result to the seventh power. This is a . Expanding all seven factors would be unnecessary work.

Name the intermediate value $u=3x-2$ . Then $y=u^7$ . The rates are $du/dx=3$ and $dy/du=7u^6$ : one measures how $u$ changes with $x$ , the other how $y$ changes with $u$ .

Multiply the two rates to obtain the rate of change of y with respect to x.

A small change in $x$ produces about three times that change in $u$ ; the change in $y$ is then about $7u^6$ times the change in $u$ . In the limit, these rates multiply. This is the chain rule:

\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}.

\begin{aligned}\frac{dy}{dx}&=7(3x-2)^6\times3\\&=21(3x-2)^6.\end{aligned}

In general, if the inside is $g(x)$ and the outside is power $n$ , keep the inside unchanged, then multiply by its derivative:

\begin{aligned}y&=[g(x)]^n,\\y'&=n[g(x)]^{n-1}g'(x).\end{aligned}

This applies where both required derivatives exist.

Differentiate the complete inside

A small positive-integer power may be easy to expand. Use the chain rule when a large, negative or fractional power makes expansion unsuitable.

Worked example

Differentiate $y=\dfrac2{(3x^2+1)^5}$ .

Rewrite as $2(3x^2+1)^{-5}$ . The outside is $2u^{-5}$ , and the inside is $u=3x^2+1$ , whose derivative is $6x$ .

\begin{aligned}\frac{dy}{dx}&=-10(3x^2+1)^{-6}\\&\quad\times6x\\&=-\frac{60x}{(3x^2+1)^6}.\end{aligned}

The denominator never vanishes because $3x^2+1>0$ for every real $x$ .

For $y=(4-\sqrt{x})^3$ , the root is inside the bracket. Write $u=4-x^{1/2}$ , so $u'=-\tfrac12x^{-1/2}$ . The minus sign stays because the root is subtracted:

y'=3(4-\sqrt{x})^2\left(-\frac1{2\sqrt{x}}\right).

The function allows $x\ge0$ , but this derivative needs $x>0$ because its denominator contains $\sqrt{x}$ .

Work with roots and fractions

Your turnIndependent practice

(a) Find the derivative of $y=\sqrt{2x^3+5}$ where $2x^3+5>0$ .

(b) Find the gradient of $y=(x+1/x)^3$ at $x=2$ . State the excluded x-value.

Show worked answer

(a) Write the square root as power $1/2$ . The inside derivative is $6x^2$ .

\begin{aligned}\frac{dy}{dx}&=\tfrac12(2x^3+5)^{-1/2}\\&\quad\times6x^2\\&=\frac{3x^2}{\sqrt{2x^3+5}}.\end{aligned}

The positive-inside condition keeps the square root real and the derivative's denominator non-zero.

(b) Write the inside as $x+x^{-1}$ ; its derivative is $1-x^{-2}$ . Keep $x\ne0$ .

y'=3(x+1/x)^2(1-x^{-2}).

At $x=2$ , the gradient is $3(5/2)^2(3/4)=225/16$ .

Choose the first step

Your turnIndependent practice

Differentiate each expression. First write your chosen method and why it suits the expression.

\begin{gathered}(x+1)(x-2),\\ (x+1)^7,\\ \frac{x+1}{x^2}.\end{gathered}

Show worked answer

Expand two short brackets: $x^2-x-2$ gives $2x-1$ .
Use the chain rule to avoid seven-factor expansion: $7(x+1)^6$ .
Split the fraction into $x^{-1}+x^{-2}$ : the derivative is $-x^{-2}-2x^{-3}$ , for $x\ne0$ .

Account for both changes

In $y=(3x-2)^7$ , first calculate $3x-2$ , then raise the result to the seventh power. This is a . Expanding all seven factors would be unnecessary work.

Name the intermediate value $u=3x-2$ . Then $y=u^7$ . The rates are $du/dx=3$ and $dy/du=7u^6$ : one measures how $u$ changes with $x$ , the other how $y$ changes with $u$ .

Multiply the two rates to obtain the rate of change of y with respect to x.

A small change in $x$ produces about three times that change in $u$ ; the change in $y$ is then about $7u^6$ times the change in $u$ . In the limit, these rates multiply. This is the chain rule:

\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}.

\begin{aligned}\frac{dy}{dx}&=7(3x-2)^6\times3\\&=21(3x-2)^6.\end{aligned}

In general, if the inside is $g(x)$ and the outside is power $n$ , keep the inside unchanged, then multiply by its derivative:

\begin{aligned}y&=[g(x)]^n,\\y'&=n[g(x)]^{n-1}g'(x).\end{aligned}

This applies where both required derivatives exist.

Differentiate the complete inside

A small positive-integer power may be easy to expand. Use the chain rule when a large, negative or fractional power makes expansion unsuitable.

Worked example

Differentiate $y=\dfrac2{(3x^2+1)^5}$ .

Rewrite as $2(3x^2+1)^{-5}$ . The outside is $2u^{-5}$ , and the inside is $u=3x^2+1$ , whose derivative is $6x$ .

\begin{aligned}\frac{dy}{dx}&=-10(3x^2+1)^{-6}\\&\quad\times6x\\&=-\frac{60x}{(3x^2+1)^6}.\end{aligned}

The denominator never vanishes because $3x^2+1>0$ for every real $x$ .

For $y=(4-\sqrt{x})^3$ , the root is inside the bracket. Write $u=4-x^{1/2}$ , so $u'=-\tfrac12x^{-1/2}$ . The minus sign stays because the root is subtracted:

y'=3(4-\sqrt{x})^2\left(-\frac1{2\sqrt{x}}\right).

The function allows $x\ge0$ , but this derivative needs $x>0$ because its denominator contains $\sqrt{x}$ .

Work with roots and fractions

Your turnIndependent practice

(a) Find the derivative of $y=\sqrt{2x^3+5}$ where $2x^3+5>0$ .

(b) Find the gradient of $y=(x+1/x)^3$ at $x=2$ . State the excluded x-value.

Show worked answer

(a) Write the square root as power $1/2$ . The inside derivative is $6x^2$ .

\begin{aligned}\frac{dy}{dx}&=\tfrac12(2x^3+5)^{-1/2}\\&\quad\times6x^2\\&=\frac{3x^2}{\sqrt{2x^3+5}}.\end{aligned}

The positive-inside condition keeps the square root real and the derivative's denominator non-zero.

(b) Write the inside as $x+x^{-1}$ ; its derivative is $1-x^{-2}$ . Keep $x\ne0$ .

y'=3(x+1/x)^2(1-x^{-2}).

At $x=2$ , the gradient is $3(5/2)^2(3/4)=225/16$ .

Choose the first step

Your turnIndependent practice

Differentiate each expression. First write your chosen method and why it suits the expression.

\begin{gathered}(x+1)(x-2),\\ (x+1)^7,\\ \frac{x+1}{x^2}.\end{gathered}

Show worked answer

Expand two short brackets: $x^2-x-2$ gives $2x-1$ .
Use the chain rule to avoid seven-factor expansion: $7(x+1)^6$ .
Split the fraction into $x^{-1}+x^{-2}$ : the derivative is $-x^{-2}-2x^{-3}$ , for $x\ne0$ .