Differentiating Roots and Fractions | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Express roots and fractions as powers

The power rule needs a coefficient multiplied by a power of $x$ . A fractional exponent represents a root; a negative exponent represents a reciprocal.

\sqrt{x}=x^{1/2},\qquad \frac1{x^2}=x^{-2}.

For a cube root, $\sqrt[3]{x^2}=x^{2/3}$ . For a root in the denominator, combine both ideas: $1/\sqrt{x}=x^{-1/2}$ .

Subtract one carefully: $-2-1=-3$ and $\tfrac12-1=-\tfrac12$ . Therefore

\begin{aligned}\frac{d}{dx}(x^{-2})&=-2x^{-3},\\ \frac{d}{dx}(x^{1/2})&=\tfrac12x^{-1/2}.\end{aligned}

The original expression still controls the domain. $1/x^2$ excludes $x=0$ . Although $\sqrt{x}$ exists at zero, its derivative $1/(2\sqrt{x})$ is finite only for $x>0$ .

Simplify products and quotients first

A quotient with a single power in the denominator can be divided term by term. Subtract exponents when dividing powers of the same base:

\frac{2x^3+5}{x^2}=2x+5x^{-2}.

Keep $x\ne0$ . Its derivative is $2-10x^{-3}$ . Do not differentiate the numerator and denominator separately and divide their derivatives.

Worked example

Find the gradient of $y=x(2x-1)(x+3)$ at $x=1$ .

Expand the two brackets, then multiply by $x$ :

\begin{aligned}y&=x(2x^2+5x-3)\\&=2x^3+5x^2-3x.\end{aligned}

Now $dy/dx=6x^2+10x-3$ . At $x=1$ , the gradient is $13$ . Expanding makes each term ready for the power rule; multiplying the derivatives of the brackets would not work.

Choose the preparation yourself

Your turnIndependent practice

(a) Differentiate $y=3x^4-\dfrac1{2x^2}+\dfrac4{\sqrt{x}}+5$ , stating its domain.

(b) Find the gradient of $xy=12$ at $x=2$ .

Show worked answer

(a) The denominator $\sqrt{x}$ requires $x>0$ . Rewrite before differentiating:

\begin{aligned}y&=3x^4-\tfrac12x^{-2}\\&\quad+4x^{-1/2}+5,\\ \frac{dy}{dx}&=12x^3+x^{-3}\\&\quad-2x^{-3/2}.\end{aligned}

The reciprocal terms come from $(-\tfrac12)(-2)=1$ and $4(-\tfrac12)=-2$ .

(b) Make $y$ the subject: $y=12/x=12x^{-1}$ , with $x\ne0$ . Then $dy/dx=-12x^{-2}$ , giving gradient $-3$ at $x=2$ .