# 3.9 Derivatives of exponential and logarithmic functions  (Page 4/6)

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## Problem-solving strategy: using logarithmic differentiation

1. To differentiate $y=h\left(x\right)$ using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain $\text{ln}\phantom{\rule{0.2em}{0ex}}y=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(h\left(x\right)\right).$
2. Use properties of logarithms to expand $\text{ln}\phantom{\rule{0.1em}{0ex}}\left(h\left(x\right)\right)$ as much as possible.
3. Differentiate both sides of the equation. On the left we will have $\frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}.$
4. Multiply both sides of the equation by $y$ to solve for $\frac{dy}{dx}.$
5. Replace $y$ by $h\left(x\right).$

## Using logarithmic differentiation

Find the derivative of $y={\left(2{x}^{4}+1\right)}^{\text{tan}\phantom{\rule{0.1em}{0ex}}x}.$

Use logarithmic differentiation to find this derivative.

$\begin{array}{cccccc}\hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{ln}{\left(2{x}^{4}+1\right)}^{\text{tan}\phantom{\rule{0.1em}{0ex}}x}\hfill & & & \text{Step 1. Take the natural logarithm of both sides.}\hfill \\ \hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{tan}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2{x}^{4}+1\right)\hfill & & & \text{Step 2. Expand using properties of logarithms.}\hfill \\ \hfill \frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}& =\hfill & {\text{sec}}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2{x}^{4}+1\right)+\frac{8{x}^{3}}{2{x}^{4}+1}·\text{tan}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \begin{array}{c}\text{Step 3. Differentiate both sides. Use the}\hfill \\ \text{product rule on the right.}\hfill \end{array}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & y·\left({\text{sec}}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2{x}^{4}+1\right)+\frac{8{x}^{3}}{2{x}^{4}+1}·\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)\hfill & & & \text{Step 4. Multiply by}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{on both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & {\left(2{x}^{4}+1\right)}^{\text{tan}\phantom{\rule{0.1em}{0ex}}x}\left({\text{sec}}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2{x}^{4}+1\right)+\frac{8{x}^{3}}{2{x}^{4}+1}·\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)\hfill & & & \text{Step 5. Substitute}\phantom{\rule{0.2em}{0ex}}y={\left(2{x}^{4}+1\right)}^{\text{tan}\phantom{\rule{0.1em}{0ex}}x}.\hfill \end{array}$

## Using logarithmic differentiation

Find the derivative of $y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}.$

This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.

$\begin{array}{cccccc}\hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{ln}\phantom{\rule{0.2em}{0ex}}\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}\hfill & & & \text{Step 1. Take the natural logarithm of both sides.}\hfill \\ \hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{ln}\phantom{\rule{0.1em}{0ex}}x+\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2x+1\right)-x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}e-3\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \text{Step 2. Expand using properties of logarithms.}\hfill \\ \hfill \frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}& =\hfill & \frac{1}{x}+\frac{1}{2x+1}-1-3\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{\text{sin}\phantom{\rule{0.1em}{0ex}}x}\hfill & & & \text{Step 3. Differentiate both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & y\left(\frac{1}{x}+\frac{1}{2x+1}-1-3\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x\right)\hfill & & & \text{Step 4. Multiply by}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{on both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & \frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}\left(\frac{1}{x}+\frac{1}{2x+1}-1-3\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x\right)\hfill & & & \text{Step 5. Substitute}\phantom{\rule{0.2em}{0ex}}y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}.\hfill \end{array}$

## Extending the power rule

Find the derivative of $y={x}^{r}$ where $r$ is an arbitrary real number.

The process is the same as in [link] , though with fewer complications.

$\begin{array}{cccccc}\hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{ln}{x}^{r}\hfill & & & \text{Step 1. Take the natural logarithm of both sides.}\hfill \\ \hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & r\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \text{Step 2. Expand using properties of logarithms.}\hfill \\ \hfill \frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}& =\hfill & r\frac{1}{x}\hfill & & & \text{Step 3. Differentiate both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & y\frac{r}{x}\hfill & & & \text{Step 4. Multiply by}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{on both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & {x}^{r}\frac{r}{x}\hfill & & & \text{Step 5. Substitute}\phantom{\rule{0.2em}{0ex}}y={x}^{r}.\hfill \\ \hfill \frac{dy}{dx}& =\hfill & r{x}^{r-1}\hfill & & & \text{Simplify.}\hfill \end{array}$

Use logarithmic differentiation to find the derivative of $y={x}^{x}.$

$\frac{dy}{dx}={x}^{x}\left(1+\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)$

Find the derivative of $y={\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)}^{\pi }.$

${y}^{\prime }=\pi {\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)}^{\pi -1}{\text{sec}}^{2}x$

## Key concepts

• On the basis of the assumption that the exponential function $y={b}^{x},b>0$ is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.
• We can use a formula to find the derivative of $y=\text{ln}\phantom{\rule{0.1em}{0ex}}x,$ and the relationship ${\text{log}}_{b}x=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{ln}\phantom{\rule{0.1em}{0ex}}b}$ allows us to extend our differentiation formulas to include logarithms with arbitrary bases.
• Logarithmic differentiation allows us to differentiate functions of the form $y=g{\left(x\right)}^{f\left(x\right)}$ or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.

## Key equations

• Derivative of the natural exponential function
$\frac{d}{dx}\left({e}^{g\left(x\right)}\right)={e}^{g\left(x\right)}{g}^{\prime }\left(x\right)$
• Derivative of the natural logarithmic function
$\frac{d}{dx}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}g\left(x\right)\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right)$
• Derivative of the general exponential function
$\frac{d}{dx}\left({b}^{g\left(x\right)}\right)={b}^{g\left(x\right)}{g}^{\prime }\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b$
• Derivative of the general logarithmic function
$\frac{d}{dx}\left({\text{log}}_{b}g\left(x\right)\right)=\frac{{g}^{\prime }\left(x\right)}{g\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}$

For the following exercises, find ${f}^{\prime }\left(x\right)$ for each function.

$f\left(x\right)={x}^{2}{e}^{x}$

$2x{e}^{x}+{x}^{2}{e}^{x}$

$f\left(x\right)=\frac{{e}^{\text{−}x}}{x}$

$f\left(x\right)={e}^{{x}^{3}\text{ln}\phantom{\rule{0.1em}{0ex}}x}$

${e}^{{x}^{3}\text{ln}\phantom{\rule{0.1em}{0ex}}x}\left(3{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}x+{x}^{2}\right)$

$f\left(x\right)=\sqrt{{e}^{2x}+2x}$

$f\left(x\right)=\frac{{e}^{x}-{e}^{\text{−}x}}{{e}^{x}+{e}^{\text{−}x}}$

$\frac{4}{{\left({e}^{x}+{e}^{\text{−}x}\right)}^{2}}$

$f\left(x\right)=\frac{{10}^{x}}{\text{ln}\phantom{\rule{0.1em}{0ex}}10}$

find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one
state and prove L hospital rule
I want to know about hospital rule
Faysal
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were$48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach \$60 billion?
what is the derivative of x= Arc sin (x)^1/2
y^2 = arcsin(x)
Pitior
x = sin (y^2)
Pitior
differentiate implicitly
Pitior
then solve for dy/dx
Pitior
thank you it was very helpful
morfling
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.