MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 37

Logarithmic differentiation

Sometimes it can simplify calculations to take the log of both sides before taking the derivative.

Find $\frac{dy}{dx}$ when $y=\frac{{(x+2)}^{\frac{5}{2}}}{{(x+6)}^{\frac{1}{2}}{(x+3)}^{\frac{7}{2}}}\text{.}$ $$\begin{array}{ccc}{\displaystyle \text{ln}\hspace{0.17em}y}& =& {\displaystyle \text{ln}\left(\frac{{(x+2)}^{\frac{5}{2}}}{{(x+6)}^{\frac{1}{2}}{(x+3)}^{\frac{7}{2}}}\right)}\\ & =& {\displaystyle \text{ln}{(x+2)}^{\frac{5}{2}}-\text{ln}{(x+6)}^{\frac{1}{2}}-\text{ln}{(x+3)}^{\frac{7}{2}}}\\ & =& {\displaystyle \frac{5}{2}\text{ln}(x+2)-\frac{1}{2}\text{ln}(x+6)-\frac{7}{2}\text{ln}(x+3)\text{.}}\end{array}$$ So, by taking the derivative with respect to $x,$ $$\frac{1}{y}\frac{dy}{dx}=\frac{5}{2}\frac{1}{(x+2)}-\frac{1}{2}\frac{1}{(x+6)}-\frac{7}{2}\frac{1}{(x+3)}\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle y(\frac{5}{2}\left(\frac{1}{x+2}\right)-\frac{1}{2}\left(\frac{1}{x+6}\right)-\frac{7}{2}\left(\frac{1}{x+3}\right))}\\ & =& {\displaystyle \frac{{(x+2)}^{52}}{{(x+6)}^{\frac{1}{2}}{(x+3)}^{\frac{7}{2}}}(\frac{5}{2}\left(\frac{1}{x+2}\right)-\frac{1}{2}\left(\frac{1}{x+6}\right)-\frac{7}{2}\left(\frac{1}{x+3}\right))\text{.}}\end{array}$$

If $x^m{y}^n={(x+y)}^{m+n}$ show that $\frac{dy}{dx}=yx\text{.}$

Take the log of both sides $$\text{ln}\left(x^m{y}^n\right)=\text{ln}\left({(x+y)}^{m+n}\right)\text{.}$$ So $$\text{ln}\hspace{0.17em}{x}^m+\text{ln}\hspace{0.17em}{y}^n=(m+n)\text{ln}(x+y)\text{.}$$ So $$m\text{ln}\hspace{0.17em}x+n\text{ln}\hspace{0.17em}y=(m+n)\text{ln}(x+y)\text{.}$$ Take the derivative with respect to $x\text{.}$ $$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=(m+n)\left(\frac{1}{x+y}\right)(1+\frac{dy}{dx})\text{.}$$ So $$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}+\left(\frac{m+n}{x+y}\right)\frac{dy}{dx}\text{.}$$ So $$(\frac{n}{y}-\frac{m+n}{x+y})\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\text{.}$$ So $$\left(\frac{nx+ny-my-ny}{y(x+y)}\right)\frac{dy}{dx}=\frac{mx+nx-mx-my}{x(x+y)}\text{.}$$ So $$\left(\frac{nx-my}{y}\right)\frac{dy}{dx}=\frac{nx-my}{x}\text{.}$$ So $$\frac{dy}{dx}=\frac{y}{x}\text{.}$$

Often the derivative can be done as usual.

Find $\frac{dy}{dx}$ when $y=a^x+e^{\text{tan}\hspace{0.17em}x}+{\left(\text{cot}\hspace{0.17em}x\right)}^{\text{cos}\hspace{0.17em}x}\text{.}$ $$\begin{array}{ccc}{\displaystyle y}& =& {\displaystyle {\left(e^{\text{ln}\hspace{0.17em}a}\right)}^x+e^{\text{tan}\hspace{0.17em}x}+{\left(e^{\text{ln}\left(\text{cot}\hspace{0.17em}x\right)}\right)}^{\text{cos}\hspace{0.17em}x}}\\ & =& {\displaystyle e^{x\text{ln}\hspace{0.17em}a}+e^{\text{tan}\hspace{0.17em}x}+e^{\text{cos}\hspace{0.17em}x\text{ln}\left(\text{cot}\hspace{0.17em}x\right)}\text{.}}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle e^{x\text{ln}\hspace{0.17em}a}\text{ln}\hspace{0.17em}a+e^{\text{tan}\hspace{0.17em}x}{\text{sec}}^{2}x+e^{\text{cos}\hspace{0.17em}x\text{ln}\left(\text{cot}\hspace{0.17em}x\right)}(\frac{\text{cos}\hspace{0.17em}x(-{\text{csc}}^{2}x)}{\text{cot}\hspace{0.17em}x}+(-\text{sin}\hspace{0.17em}x)\text{ln}\left(\text{cot}\hspace{0.17em}x\right))}\\ & =& {\displaystyle a^x\text{ln}\hspace{0.17em}a+e^{\text{tan}\hspace{0.17em}x}{\text{sec}}^{2}x+{\left(\text{cot}\hspace{0.17em}x\right)}^{\text{cos}\hspace{0.17em}x}(\frac{\text{cos}\hspace{0.17em}x\frac{-1}{{\text{sin}}^{2}x}}{\frac{\text{cos}\hspace{0.17em}x}{\text{sin}\hspace{0.17em}x}}+(-\text{sin}\hspace{0.17em}x)\text{ln}\left(\text{cot}\hspace{0.17em}x\right))}\\ & =& {\displaystyle a^x\text{ln}\hspace{0.17em}a+e^{\text{tan}\hspace{0.17em}x}{\text{sec}}^{2}x+{\left(\text{cot}\hspace{0.17em}x\right)}^{\text{cos}\hspace{0.17em}x}(-\text{csc}\hspace{0.17em}x-\text{sin}\hspace{0.17em}x\text{ln}\left(\text{cot}\hspace{0.17em}x\right))\text{.}}\end{array}$$

Find $\frac{dy}{dx}$ when $y=x^{x^{x^{⋰}}}\text{.}$

Then $y=x^y\text{.}$ So $y={\left(e^{\text{ln}\hspace{0.17em}x}\right)}^y=e^{y\text{ln}\hspace{0.17em}x}\text{.}$ So $$\frac{dy}{dx}=e^{y\text{ln}\hspace{0.17em}x}\frac{d\left(y\text{ln}\hspace{0.17em}x\right)}{dx}=e^{y\text{ln}\hspace{0.17em}x}(\frac{y}{x}+\frac{dy}{dx}\text{ln}\hspace{0.17em}x)\text{.}$$ So $$\frac{dy}{dx}=\frac{y}{x}{e}^{y\text{ln}\hspace{0.17em}x}+\text{ln}\hspace{0.17em}x{e}^{y\text{ln}\hspace{0.17em}x}\frac{dy}{dx}\text{.}$$ So $$(1-\text{ln}\hspace{0.17em}x{e}^{y\text{ln}\hspace{0.17em}x})\frac{dy}{dx}=\frac{y}{x}{x}^y\text{.}$$ So $$\frac{dy}{dx}=\frac{\frac{y{x}^y}{x}}{1-\text{ln}\hspace{0.17em}x·x^y}=\frac{y{x}^y}{x(1-x^y\text{ln}\hspace{0.17em}x)}\text{.}$$

Motion

The main point is: If $s=\text{distance}$ traveled then $\frac{ds}{dt}=\text{speed.}$ Speed is the same thing as velocity and $$\frac{d\hspace{0.17em}\text{speed}}{dt}=\frac{d\hspace{0.17em}\text{velocity}}{dt}=\text{acceleration.}$$

A particle falls from the top of a tower and in the last second before it hits the ground it falls $\frac{9}{25}$ of the total height of the tower. Find the height of the tower.

Tower: $$\begin{array}{c} { H \end{array}$$ Position of particle is ???
Acceleration of particle is $-9.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\text{.}$
So $$\frac{dv}{dt}=-9.8\text{.}$$ So $$dv=-9.8dt\text{.}$$ So $$\int dv=\int -9.8dt\text{.}$$ So $$v=-9.8t+c\text{.}$$ Now $v$ at time $0$ is $0\text{.}$ So $$0=-9.8·0+c\text{.}$$ So $c=0$ and $$v=-9.8t\text{.}$$ Since $v=\frac{ds}{dt},$ $$ds=vdt=-9.8tdt\text{.}$$ So $$\int ds=\int 09.8tdt\text{.}$$ So $$s=-9.8\frac{t^2}{2}+c_1\text{.}$$ At time $0,$ $s=H\text{.}$ So $$H=-\frac{9.8}{2}·0^2+c_1\text{.}$$ So $c_1=H\text{.}$ So $$s=-\frac{9.8t^2}{2}+H\text{.}$$ The particle hits the ground when $s=0\text{.}$ $$0=-\frac{9.8}{2}{t}^2+H\text{.}$$ So $$\frac{9.8}{2}{t}^2=H\text{.}$$ So $$t^2=\frac{2H}{9.8}=\frac{H}{4.9}\text{.}$$ So $$t=\sqrt{\frac{H}{4.9}}\text{.}$$ So the particle hits the ground when $t=\sqrt{\frac{H}{4.9}}\text{.}$ One second before the particle hits the ground its height is $\frac{9}{25}H\text{.}$ So when $t=\sqrt{\frac{H}{4.9}}-1,$ $s=\frac{9}{25}H\text{.}$ So $$\frac{9}{25}H=-\frac{9.8}{2}{(\sqrt{\frac{H}{4.9}}-1)}^2+H\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \frac{9}{25}H}& =& {\displaystyle -4.9(\frac{H}{4.9}-\frac{2}{\sqrt{4.9}}\sqrt{H}+1)+H}\\ & =& {\displaystyle -H+2\sqrt{4.9}\sqrt{H}-4.9+H}\\ & =& {\displaystyle 2\sqrt{4.9}\sqrt{H}-4.9\text{.}}\end{array}$$ So $$\frac{9}{25}H-2\sqrt{4.9}\sqrt{H}+4.9=0\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \sqrt{H}}& =& {\displaystyle \frac{2\sqrt{4.9}\pm \sqrt{{\left(2\sqrt{4.9}\right)}^2-\frac{4.9}{25}\left(4.9\right)}}{\frac{2.9}{25}}}\\ & =& {\displaystyle \frac{2\sqrt{4.9}\pm \sqrt{4\left(4.9\right)-4\left(4.9\right)\frac{9}{25}}}{\frac{18}{25}}}\\ & =& {\displaystyle \frac{2\sqrt{4.9}\pm 2\sqrt{4.9}\sqrt{\frac{16}{25}}}{\frac{18}{25}}}\\ & =& {\displaystyle \frac{2\sqrt{4.9}(1\pm \frac{4}{5})}{\frac{18}{25}}}\\ & =& {\displaystyle \{\begin{array}{c}\frac{\sqrt{4.9}\frac{9}{5}}{\frac{9}{25}}\\ \frac{\sqrt{4.9}\frac{1}{5}}{\frac{9}{25}}\end{array}}\\ & =& {\displaystyle \{\begin{array}{c}5\sqrt{4.9}\\ \text{or}\\ \frac{5}{9}\sqrt{4.9}\end{array}\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 37 from a series of handwritten lecture notes for the class MATH 221 given on December 8, 2000.

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