H2 Math Differentiation Techniques: Study Guide

JC · H2 Mathematics · Study guide

Work through derivatives, implicit stationary points, logarithmic differentiation, parametric tangents, a midpoint locus and a further normal intersection.

Concepts · worked methods · graphs10 min read · practice linked below

Choose the rule before differentiating

Read the whole expression first. A product needs two differentiated terms; a composite needs the derivative of its inside. If both the base and exponent vary, take logarithms. For a curve given in terms of a parameter, differentiate both coordinates with respect to that parameter.

Product, quotient and chain rules

Differentiate a product, then simplify

Find the gradient of $y=x^2e^{-3x}$ at $x=1$ , and determine where its tangent is horizontal.

The exponential contains an inner function, so its derivative includes a factor of −3. Use the product rule on the whole expression.

\begin{aligned}y'&=2xe^{-3x}+x^2(-3e^{-3x})\\&=xe^{-3x}(2-3x).\\y'(1)&=-e^{-3}.\end{aligned}

A horizontal tangent requires $y'=0$ . Since the exponential is positive,

\begin{aligned}x(2-3x)&=0,\\x&=0\quad\text{or}\quad x=\frac23.\end{aligned}

Factorising the derivative makes the horizontal-tangent condition much easier to solve.

Differentiate a quotient containing a logarithm

Find $y'$ for $y=\dfrac{\ln x}{1+x^2}$ , $x>0$ . Keep the denominator squared until the numerator is complete.

\begin{aligned}y'&=\frac{(1+x^2)/x-2x\ln x}{(1+x^2)^2}\\&=\frac{1+x^2-2x^2\ln x}{x(1+x^2)^2}.\end{aligned}

At $x=1$ , the gradient is $1/2$ . The domain $x>0$ comes from the original logarithm.

An inverse trigonometric function is a composite

Find the derivative of $y=\cos^{-1}(x^{3/2})$ in non-trigonometric form.

The real-valued function has domain $0\leq x\leq1$ . On $0<x<1$ , use $u=x^{3/2}$ :

\begin{aligned}\frac{dy}{dx}&=-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}\\&=-\frac{1}{\sqrt{1-x^3}}\left(\frac32\sqrt{x}\right)\\&=-\frac{3\sqrt{x}}{2\sqrt{1-x^3}}.\end{aligned}

Here $\cos^{-1}$ means inverse cosine. At $x=1$ the finite derivative formula fails; do not substitute an endpoint into a formula derived for interior points.

Implicit and logarithmic differentiation

Find stationary points on an implicit curve

The curve is $x^2+xy+y^2=3$ . Find its stationary points.

Treat $y$ as a function of $x$ . In particular, the product $xy$ differentiates to $y+xy'$ .

\begin{aligned}2x+y+xy'+2yy'&=0,\\(x+2y)y'&=-2x-y,\\y'&=-\frac{2x+y}{x+2y}.\end{aligned}

At a stationary point, $2x+y=0$ , provided $x+2y\ne0$ . Substitute $y=-2x$ into the original curve:

\begin{aligned}x^2-2x^2+4x^2&=3,\\3x^2&=3,\\x&=\pm1.\end{aligned}

The stationary points are $(1,-2)$ and $(-1,2)$ . Their derivative denominators are −3 and 3, so both tangents are horizontal. A zero numerator alone is insufficient if the denominator also vanishes.

Both the base and the power vary

Differentiate $y=(1+x^2)^x$ . Since $1+x^2>0$ , logarithms are valid for all real $x$ .

\begin{aligned}\ln y&=x\ln(1+x^2),\\\frac{y'}{y}&=\ln(1+x^2)+x\frac{2x}{1+x^2},\\y'&=(1+x^2)^x\\&\quad\times\left[\ln(1+x^2)+\frac{2x^2}{1+x^2}\right].\end{aligned}

Check: $y'(0)=0$ . Do not use the ordinary power rule while pretending that the exponent $x$ is constant.

Parametric differentiation and a midpoint locus

Find a tangent, its intercepts and the locus

A curve is given by $x=4\cos t$ , $y=3\sin t$ , where $0<t<\pi/2$ . At parameter $p$ , its tangent cuts the axes at $A$ and $B$ . Find the tangent and the Cartesian equation traced by their midpoint $M$ .

\begin{aligned}\frac{dx}{dt}&=-4\sin t,\\\frac{dy}{dt}&=3\cos t,\\\frac{dy}{dx}&=-\frac34\cot t.\end{aligned}

The tangent at $P=(4\cos p,3\sin p)$ is

\begin{aligned}y-3\sin p&=-\frac34\cot p(x-4\cos p),\\y&=-\frac34\cot p\,x\\&\quad+3\frac{\sin^2p+\cos^2p}{\sin p},\\y&=-\frac34\cot p\,x+3\csc p.\end{aligned}

Find the intercepts by setting one coordinate to zero, then average the coordinates.

\begin{aligned}A&=(4\sec p,0),\\B&=(0,3\csc p),\\M&=\left(2\sec p,\frac32\csc p\right).\end{aligned}

Writing the coordinates of $M$ as $(X,Y)$ , eliminate $p$ :

\begin{aligned}\cos p&=\frac2X,\qquad\sin p=\frac3{2Y},\\\frac4{X^2}+\frac9{4Y^2}&=1,\\X&>2,\qquad Y>\frac32.\end{aligned}

The restrictions select the branch actually traced. They are part of the locus answer.

For a second derivative, differentiate the first gradient with respect to the parameter, then divide by dx/dt again:

\begin{aligned}\frac{d^2y}{dx^2}&=\frac{\frac{d}{dt}(-\frac34\cot t)}{-4\sin t}\\&=-\frac3{16\sin^3t}<0.\end{aligned}

Watch the tangent and its intercepts

This is the same ellipse and tangent as the worked application. Change $p$ within the first quadrant. As the tangent becomes flatter, its x-intercept moves farther away.

p = 45°: A = (5.657, 0), B = (0, 4.243). The midpoint is (2.828, 2.121).

The graph shows the nearby tangent segment; the status gives both intercepts even if an intercept is outside the viewing window.

A normal can meet the curve again

Substitute the line back into the parametric equations

For $x=\frac12e^t-2e^{-t}$ , $y=e^t+2e^{-t}$ , find the normal at $t=\ln4$ , then its other intersection with the curve.

\begin{aligned}\frac{dx}{dt}&=\frac12e^t+2e^{-t},\\\frac{dy}{dt}&=e^t-2e^{-t},\\m_{\rm tangent}&=\frac{4-1/2}{2+1/2}=\frac75,\\m_{\rm normal}&=-\frac57.\end{aligned}

The point is $(3/2,9/2)$ , so

\begin{aligned}y-\frac92&=-\frac57\left(x-\frac32\right),\\5x+7y&=39.\end{aligned}

Substitute the parameter expressions into this line. Put $u=e^t>0$ to obtain a quadratic:

\begin{aligned}\frac{19}{2}e^t+4e^{-t}&=39,\\19u^2-78u+8&=0,\\(u-4)(19u-2)&=0,\\u&=4\quad\text{or}\quad\frac2{19}.\end{aligned}

The original point has $u=4$ , so the other intersection has $t=\ln(2/19)$ . Both roots are positive and both give real parameter values. Substituting the second parameter back into the coordinates gives the other intersection:

\begin{aligned}x&=\frac12\left(\frac2{19}\right)-2\left(\frac{19}{2}\right),\\&=-\frac{360}{19},\\y&=\frac2{19}+2\left(\frac{19}{2}\right),\\&=\frac{363}{19}.\end{aligned}

Hence the point is $(-360/19,363/19)$ . Checking in the normal gives $5x+7y=(-1800+2541)/19=39$ .

Horizontal and vertical tangents

For $x=t^2$ , $y=t$ , the tangent direction is $(2t,1)$ . At $t=0$ , dx/dt is zero but dy/dt is not: the tangent is the vertical line $x=0$ , and the normal is $y=0$ . For a horizontal tangent, dy/dt must be zero while dx/dt is non-zero. If both vanish, examine the curve or a limiting direction; neither shortcut settles it.

Before you finish

Check the original domain, evaluate the point as well as the gradient, and reject the known intersection only after factoring. For a locus, retain the parameter restrictions.

Differentiation practice questions and worked solutions