# 3.9 Derivatives of exponential and logarithmic functions  (Page 3/6)

 Page 3 / 6

## Proof

If $x>0$ and $y=\text{ln}\phantom{\rule{0.1em}{0ex}}x,$ then ${e}^{y}=x.$ Differentiating both sides of this equation results in the equation

${e}^{y}\frac{dy}{dx}=1.$

Solving for $\frac{dy}{dx}$ yields

$\frac{dy}{dx}=\frac{1}{{e}^{y}}.$

Finally, we substitute $x={e}^{y}$ to obtain

$\frac{dy}{dx}=\frac{1}{x}.$

We may also derive this result by applying the inverse function theorem, as follows. Since $y=g\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ is the inverse of $f\left(x\right)={e}^{x},$ by applying the inverse function theorem we have

$\frac{dy}{dx}=\frac{1}{{f}^{\prime }\left(g\left(x\right)\right)}=\frac{1}{{e}^{\text{ln}\phantom{\rule{0.1em}{0ex}}x}}=\frac{1}{x}.$

Using this result and applying the chain rule to $h\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right)$ yields

${h}^{\prime }\left(x\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right).$

The graph of $y=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ and its derivative $\frac{dy}{dx}=\frac{1}{x}$ are shown in [link] .

## Taking a derivative of a natural logarithm

Find the derivative of $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}+3x-4\right).$

$\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\frac{1}{{x}^{3}+3x-4}·\left(3{x}^{2}+3\right)\hfill & & & \text{Use}\phantom{\rule{0.2em}{0ex}}g\left(x\right)={x}^{3}+3x-4\phantom{\rule{0.2em}{0ex}}\text{in}\phantom{\rule{0.2em}{0ex}}{h}^{\prime }\left(x\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right).\hfill \\ & =\frac{3{x}^{2}+3}{{x}^{3}+3x-4}\hfill & & & \text{Rewrite.}\hfill \end{array}$

## Using properties of logarithms in a derivative

Find the derivative of $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\frac{{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x}{2x+1}\right).$

At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.

$\begin{array}{cccccc}\hfill f\left(x\right)& =\hfill & \text{ln}\phantom{\rule{0.1em}{0ex}}\left(\frac{{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x}{2x+1}\right)=2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x+\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)-\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2x+1\right)\hfill & & & \text{Apply properties of logarithms.}\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & \frac{2}{x}+\text{cot}\phantom{\rule{0.1em}{0ex}}x-\frac{2}{2x+1}\hfill & & & \text{Apply sum rule and}\phantom{\rule{0.2em}{0ex}}{h}^{\prime }\left(x\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right).\hfill \end{array}$

Differentiate: $f\left(x\right)=\text{ln}{\left(3x+2\right)}^{5}.$

${f}^{\prime }\left(x\right)=\frac{15}{3x+2}$

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of $y=lo{g}_{b}x$ and $y={b}^{x}$ for $b>0,b\ne 1.$

## Derivatives of general exponential and logarithmic functions

Let $b>0,b\ne 1,$ and let $g\left(x\right)$ be a differentiable function.

1. If, $y={\text{log}}_{b}x,$ then
$\frac{dy}{dx}=\frac{1}{x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}.$

More generally, if $h\left(x\right)={\text{log}}_{b}\left(g\left(x\right)\right),$ then for all values of x for which $g\left(x\right)>0,$
${h}^{\prime }\left(x\right)=\frac{{g}^{\prime }\left(x\right)}{g\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}.$
2. If $y={b}^{x},$ then
$\frac{dy}{dx}={b}^{x}\text{ln}\phantom{\rule{0.1em}{0ex}}b.$

More generally, if $h\left(x\right)={b}^{g\left(x\right)},$ then
${h}^{\prime }\left(x\right)={b}^{g\left(x\right)}g\text{″}\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b.$

## Proof

If $y={\text{log}}_{b}x,$ then ${b}^{y}=x.$ It follows that $\text{ln}\phantom{\rule{0.1em}{0ex}}\left({b}^{y}\right)=\text{ln}\phantom{\rule{0.2em}{0ex}}x.$ Thus $y\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}b=\text{ln}\phantom{\rule{0.2em}{0ex}}x.$ Solving for $y,$ we have $y=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{ln}\phantom{\rule{0.1em}{0ex}}b}.$ Differentiating and keeping in mind that $\text{ln}\phantom{\rule{0.1em}{0ex}}b$ is a constant, we see that

$\frac{dy}{dx}=\frac{1}{x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}.$

The derivative in [link] now follows from the chain rule.

If $y={b}^{x},$ then $\text{ln}\phantom{\rule{0.2em}{0ex}}y=x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b.$ Using implicit differentiation, again keeping in mind that $\text{ln}\phantom{\rule{0.1em}{0ex}}b$ is constant, it follows that $\frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}=\text{ln}\phantom{\rule{0.1em}{0ex}}b.$ Solving for $\frac{dy}{dx}$ and substituting $y={b}^{x},$ we see that

$\frac{dy}{dx}=y\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b={b}^{x}\text{ln}\phantom{\rule{0.1em}{0ex}}b.$

The more general derivative ( [link] ) follows from the chain rule.

## Applying derivative formulas

Find the derivative of $h\left(x\right)=\frac{{3}^{x}}{{3}^{x}+2}.$

Use the quotient rule and [link] .

$\begin{array}{ccccc}\hfill {h}^{\prime }\left(x\right)& =\frac{{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3\left({3}^{x}+2\right)-{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3\left({3}^{x}\right)}{{\left({3}^{x}+2\right)}^{2}}\hfill & & & \text{Apply the quotient rule.}\hfill \\ & =\frac{2·{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3}{{\left({3}^{x}+2\right)}^{2}}\hfill & & & \text{Simplify.}\hfill \end{array}$

## Finding the slope of a tangent line

Find the slope of the line tangent to the graph of $y={\text{log}}_{2}\left(3x+1\right)$ at $x=1.$

To find the slope, we must evaluate $\frac{dy}{dx}$ at $x=1.$ Using [link] , we see that

$\frac{dy}{dx}=\frac{3}{\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2\left(3x+1\right)}.$

By evaluating the derivative at $x=1,$ we see that the tangent line has slope

$\frac{dy}{dx}|\begin{array}{l}\\ {}_{x=1}\end{array}=\frac{3}{4\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2}=\frac{3}{\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}16}.$

Find the slope for the line tangent to $y={3}^{x}$ at $x=2.$

$9\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(3\right)$

## Logarithmic differentiation

At this point, we can take derivatives of functions of the form $y={\left(g\left(x\right)\right)}^{n}$ for certain values of $n,$ as well as functions of the form $y={b}^{g\left(x\right)},$ where $b>0$ and $b\ne 1.$ Unfortunately, we still do not know the derivatives of functions such as $y={x}^{x}$ or $y={x}^{\pi }.$ These functions require a technique called logarithmic differentiation    , which allows us to differentiate any function of the form $h\left(x\right)=g{\left(x\right)}^{f\left(x\right)}.$ It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of $y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}.$ We outline this technique in the following problem-solving strategy.

Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
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list the basic elementary differentials
Differentiation and integration
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Damien
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the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
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calculus is math that studies the change in math, such as the rate and distance,
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the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
by using integration product formula
Roha
find derivative f(x)=1/x
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
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