Differentiation and its applications – university Year 1 maths lesson to download as PDF

Example 1: first principles

Differentiate \(f(x)=x^3\) from first principles.

\(\dfrac{(x+h)^3-x^3}{h}=3x^2+3xh+h^2\to3x^2\) as \(h\to0\).

Example 2

Differentiate \(y=x^x\), and find \(y'\) on \(x^2+y^2=25\).

\(\ln y=x\ln x\Rightarrow y'/y=\ln x+1\), so \(y'=x^x(\ln x+1)\). For the circle, \(2x+2y\,y'=0\), so \(y'=-x/y\).

Example 3

Find and classify the stationary points of \(f(x)=x^3-3x\).

\(f'(x)=3x^2-3=0\) gives \(x=\pm1\). Since \(f''(x)=6x\), \(f''(1)=6>0\): local minimum, \(f(1)=-2\); and \(f''(-1)=-6<0\): local maximum, \(f(-1)=2\).

Example 4: optimisation

Among rectangles of perimeter 20, find the greatest area.

Sides \(x\) and \(10-x\): \(A=x(10-x)\), \(A'=10-2x=0\) at \(x=5\), and \(A''=-2<0\). The maximum area is 25, a square.

Example 5

Evaluate \(\lim_{x\to0}\dfrac{e^x-1-x}{x^2}\) and \(\lim_{x\to0^+}x\ln x\).

First: \(\lim\frac{e^x-1}{2x}=\lim\frac{e^x}{2}=\frac12\). Second: write \(x\ln x=\frac{\ln x}{1/x}\), which gives \(\frac{1/x}{-1/x^2}=-x\to0\).

Example 6

Estimate \(e^{0.1}\).

\(e^{0.1}\approx1+0.1+0.005+0.000167=1.105167\); the true value is \(1.105171\).

Worked example 7: The quotient rule with simplification

Differentiate \(y=\dfrac{\ln x}{x^2}\) and find where \(y'=0\).

Take \(u=\ln x\) and \(v=x^2\). Then \(y'=\dfrac{\frac1x\cdot x^2-\ln x\cdot2x}{x^4}=\dfrac{x-2x\ln x}{x^4}=\dfrac{1-2\ln x}{x^3}\). Setting the numerator to zero gives \(\ln x=\tfrac12\), so \(x=\sqrt e\approx1.65\). Because \(y'\) is positive before this point and negative after it, the function has a maximum there, with value \(\dfrac{1/2}{e}=\dfrac1{2e}\approx0.184\).

Worked example 8: A tangent to an implicit curve

Find the tangent to \(x^2+xy+y^2=7\) at \((2,1)\).

Differentiating implicitly gives \(2x+y+x\,y'+2y\,y'=0\). At \((2,1)\): \(4+1+2y'+2y'=0\), so \(y'=-\tfrac54\). The tangent is \(y-1=-\tfrac54(x-2)\), that is \(5x+4y=14\). As a check, the point lies on the curve because \(4+2+1=7\). Notice that we never needed to solve for \(y\) in terms of \(x\).

Worked example 9: Using the mean value theorem to prove an inequality

Prove that \(\ln(1+x)\lt x\) for \(x>0\).

Apply the mean value theorem to \(f(t)=\ln(1+t)\) on \([0,x]\). It is continuous and differentiable, so there is a number \(c\) between \(0\) and \(x\) with \(\dfrac{\ln(1+x)-\ln1}{x}=\dfrac1{1+c}\). Since \(c>0\) we have \(\dfrac1{1+c}<1\), and so \(\ln(1+x)\lt x\). The same method proves many inequalities: choose a function, apply the theorem on a suitable interval, and bound the derivative at the unknown point.

Worked example 10: Composite Maclaurin series

Find the Maclaurin series of \(\dfrac{\sin x}{x}\) up to \(x^4\) and use it to evaluate \(\lim_{x\to0}\dfrac{\sin x}x\).

Divide the series for \(\sin x\) by \(x\): \(\dfrac{\sin x}x=1-\dfrac{x^2}6+\dfrac{x^4}{120}-\cdots\). As \(x\to0\) every term except the first vanishes, so the limit is \(1\). This is more informative than l'Hôpital's rule because it also tells us that near zero \(\dfrac{\sin x}x\approx1-\dfrac{x^2}6\), so the function dips below 1 as \(x\) moves away from 0.

Worked example 11: A limit of the form infinity minus infinity

Evaluate \(\displaystyle\lim_{x\to0}\left(\dfrac1x-\dfrac1{\sin x}\right)\).

Combine the fractions: \(\dfrac{\sin x-x}{x\sin x}\), which has the form \(\dfrac00\). Differentiating top and bottom gives \(\dfrac{\cos x-1}{\sin x+x\cos x}\), still \(\dfrac00\). Once more: \(\dfrac{-\sin x}{2\cos x-x\sin x}\to\dfrac02=0\). So the limit is \(0\). Rewriting the expression so that it is a single quotient is the key step whenever you meet \(\infty-\infty\).

Worked example 12: The inverse function rule

Show that \(\dfrac d{dx}\arctan x=\dfrac1{1+x^2}\).

Let \(y=\arctan x\), so \(x=\tan y\). Then \(\dfrac{dx}{dy}=\sec^2y=1+\tan^2y=1+x^2\). The inverse function rule says that \(\dfrac{dy}{dx}=\dfrac1{dx/dy}\), so \(\dfrac{dy}{dx}=\dfrac1{1+x^2}\). The same idea gives the derivatives of \(\arcsin x\) and \(\ln x\) from those of \(\sin y\) and \(e^y\).

Worked example 13: Concavity and the second derivative

Find the intervals on which \(f(x)=x^4-6x^2\) is concave up, and find its points of inflection.

The first derivative is \(f'(x)=4x^3-12x\) and the second is \(f''(x)=12x^2-12=12(x-1)(x+1)\). The second derivative is positive when \(|x|>1\), so \(f\) is concave up on \((-\infty,-1)\) and \((1,\infty)\), and negative on \((-1,1)\), where \(f\) is concave down. The sign of \(f''\) changes at \(x=\pm1\), so these are points of inflection, at \((\pm1,-5)\). A change of curvature is the signature of an inflection point.

Worked example 14: Estimating with a linear approximation

Use a tangent line to estimate \(\sqrt{101}\) and state whether the estimate is too large or too small.

Take \(f(x)=\sqrt x\) near \(a=100\). Then \(f(100)=10\) and \(f'(100)=\dfrac1{2\sqrt{100}}=0.05\), so \(\sqrt{101}\approx10+0.05\times1=10.05\). Since \(f''(x)=-\tfrac14x^{-3/2}<0\) the graph is concave down, so the tangent lies above the curve and the estimate is slightly too large. The true value is \(10.04988\), and the error of about \(0.0001\) is just what the second-order term \(-\dfrac{1}{8000}\) predicts.

Worked example 15: Differentiability at a point

Is \(f(x)=x|x|\) differentiable at \(x=0\)? Use the definition.

Compute the difference quotient at \(0\): \(\dfrac{f(h)-f(0)}h=\dfrac{h|h|}h=|h|\). As \(h\to0\) from either side, \(|h|\to0\), so the limit exists and \(f'(0)=0\). The function is therefore differentiable at \(0\), even though it is defined using the absolute value. Compare this with \(|x|\) itself, whose difference quotient is \(\dfrac{|h|}h\), equal to \(1\) for positive \(h\) and \(-1\) for negative \(h\), so no limit exists. Checking the one-sided limits is the safest way to decide.