Second Derivatives | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Differentiate the gradient function

The first derivative is itself a function of $x$ . Differentiate it again to measure how the gradient changes as $x$ changes. The result is the , written $f''(x)$ or $\dfrac{d^2y}{dx^2}$ .

The superscript $2$ tells you to differentiate twice. It does not mean square the first derivative.

\begin{aligned}f(x)&=x^2-4x,\\ f'(x)&=2x-4,\\ f''(x)&=2.\end{aligned}

A negative gradient can be increasing: it becomes less negative, reaches zero, then becomes positive.

Your turnIndependent practice

At one point, $f'(x)=-3$ and $f''(x)=2$ . As $x$ increases here, does the curve rise or fall? Is its gradient increasing or decreasing?

Show worked answer

The curve falls because its gradient is negative. The gradient is increasing because the second derivative is positive: it is becoming less negative, not necessarily positive.

Apply the rules again

Worked example

Find the second derivative of $y=5(2x-3)^{-3}$ .

The inside derivative is $2$ on both differentiations. The first coefficient is $5(-3)(2)=-30$ ; the next is $(-30)(-4)(2)=240$ . Keep $x\ne3/2$ .

\begin{aligned}\frac{dy}{dx}&=-30(2x-3)^{-4},\\ \frac{d^2y}{dx^2}&=240(2x-3)^{-5}.\end{aligned}

If the question already gives $f'(x)$ , differentiate that expression only once to get $f''(x)$ . Substitute a requested x-coordinate after the differentiation.

Keep the derivative order clear

Your turn9709/12 May/Jun 2020 Q10(a) [4 marks]

For $y=54x-(2x-7)^3$ , find $dy/dx$ and $d^2y/dx^2$ .

Show worked answer

\begin{aligned}\frac{dy}{dx}&=54-6(2x-7)^2,\\ \frac{d^2y}{dx^2}&=-24(2x-7).\end{aligned}

The first chain factor is $3\times2=6$ ; the second is $(-6)\times2\times2=-24$ . The constant $54$ disappears in the second differentiation. Expanding the brackets first gives the same derivatives and provides a check.

Your turnIndependent practice

Given $f'(x)=3(2x-1)^{-8}$ , find $f''(1)$ . Do you need to find $f(x)$ first?

Show worked answer

No. Differentiate the given first derivative: $f''(x)=-48(2x-1)^{-9}$ . Hence $f''(1)=-48$ .

Differentiate the gradient function

The first derivative is itself a function of $x$ . Differentiate it again to measure how the gradient changes as $x$ changes. The result is the , written $f''(x)$ or $\dfrac{d^2y}{dx^2}$ .

The superscript $2$ tells you to differentiate twice. It does not mean square the first derivative.

\begin{aligned}f(x)&=x^2-4x,\\ f'(x)&=2x-4,\\ f''(x)&=2.\end{aligned}

A negative gradient can be increasing: it becomes less negative, reaches zero, then becomes positive.

Your turnIndependent practice

At one point, $f'(x)=-3$ and $f''(x)=2$ . As $x$ increases here, does the curve rise or fall? Is its gradient increasing or decreasing?

Show worked answer

The curve falls because its gradient is negative. The gradient is increasing because the second derivative is positive: it is becoming less negative, not necessarily positive.

Apply the rules again

Worked example

Find the second derivative of $y=5(2x-3)^{-3}$ .

The inside derivative is $2$ on both differentiations. The first coefficient is $5(-3)(2)=-30$ ; the next is $(-30)(-4)(2)=240$ . Keep $x\ne3/2$ .

\begin{aligned}\frac{dy}{dx}&=-30(2x-3)^{-4},\\ \frac{d^2y}{dx^2}&=240(2x-3)^{-5}.\end{aligned}

If the question already gives $f'(x)$ , differentiate that expression only once to get $f''(x)$ . Substitute a requested x-coordinate after the differentiation.

Keep the derivative order clear

Your turn9709/12 May/Jun 2020 Q10(a) [4 marks]

For $y=54x-(2x-7)^3$ , find $dy/dx$ and $d^2y/dx^2$ .

Show worked answer

\begin{aligned}\frac{dy}{dx}&=54-6(2x-7)^2,\\ \frac{d^2y}{dx^2}&=-24(2x-7).\end{aligned}

Your turnIndependent practice

Given $f'(x)=3(2x-1)^{-8}$ , find $f''(1)$ . Do you need to find $f(x)$ first?

Show worked answer

No. Differentiate the given first derivative: $f''(x)=-48(2x-1)^{-9}$ . Hence $f''(1)=-48$ .