4.4 Logarithmic differentiation

Each speaker produces 3dB of noise, so if $P$ is the power of one speaker, $3=V(P)=10\log_{10}\left(\frac{P}{S}\right)$. Use this to find $V(10P)$ and $V(100P)$.

Ten speakers: 13 dB. One hundred speakers: 23 dB.

We are given that one speaker produces 3dB. So if $P$ is the power of one speaker,

So, for ten speakers:

and for one hundred speakers:

The question asks you when $A(t)=2000$. So, solve $2000=1000e^{t/20}$ for $t$.

$20\log2 \approx 14$ years

The investment doubles when it hits $2000. So, we find the value of $t$ that gives $A(t)=2000$:

What happens when $\cos x$ is a negative number?

(b)

From our logarithm rules, we know that when $y$ is positive, $\log (y^2)=2\log y$. However, the expression $\cos x$ does not always take on positive values, so (a) is not correct. (For instance, when $x=\pi$, $\log(\cos^2 x)=\log(\cos^2\pi)=\log\left((-1)^2\right) = \log(1)=0$, while $2\log (\cos \pi)=2\log(-1)$, which does not exist.)

Because $\cos^2 x$ is never negative, we notice that $\cos^2 x = |\cos x|^2$. When $\cos x$ is nonzero, $|\cos x|$ is positive, so our logarithm rules tell us $\log\left(|\cos x|^2\right) =2\log|\cos x |$. When $\cos x$ is exactly zero, then both $\log(\cos^2x)$ and $2\log|\cos x|$ do not exist. So, $\log(\cos^2x) = 2\log|\cos x|$.

There are two easy ways: use the chain rule, or simplify first.

$f'(x)=\dfrac{1}{x}$

There are two easy ways: use the chain rule, or simplify first.

$f'(x)=\dfrac{2}{x}$

Don't be fooled by a common mistake: $\log(x^2+x)$ is not the same as $\log(x^2)+\log x$.

$f'(x)=\dfrac{2x+1}{x^2+x}$

Don't be fooled by a common mistake: $\log(x^2+x)$ is not the same as $\log(x^2)+\log x$.
We differentiate using the chain rule: $\ds\diff{}{x}\left\{\log(x^2+x)\right\}=\dfrac{1}{x^2+x}\cdot(2x+1) = \dfrac{2x+1}{x^2+x}$.

Use the base-change formula to convert this to natural logarithm (base $e$).

$f'(x) = \dfrac{1}{x\log10}$

We know the derivative of the natural logarithm (base $e$), so we use the base-change formula:

Since $\log 10$ is a constant:

$y'=\dfrac{1-3\log x}{x^4}$

Use the chain rule.

$\ds\diff{}{\theta} \log(\sec \theta) = \tan \theta$

Using the chain rule,

Remark: the domain of the function $\log(\sec \theta)$ is those values of $\theta$ for which $\sec\theta$ is positive: so, the intervals $\left(\left(2n-\frac{1}{2}\right)\pi,\left(2n+\frac{1}{2}\right)\pi\right)$ where $n$ is any integer. Certainly the tangent function has a larger domain than this, but outside the domain of $\log(\sec \theta)$, $\tan \theta$ is not the derivative of $\log(\sec \theta)$.

Use the chain rule twice.

$f'(x)=\dfrac{-e^{\cos(\log x)}\sin(\log x)}{x}$

Let's start in with the chain rule.

We'll need the chain rule again:

Remark: Although we have a logarithm in the exponent, we can't cancel. The expression $e^{\cos (\log x)}$ is not the same as the expression $x^{\cos x}$, or $\cos x$.

You'll need to use the chain rule twice.

$y'=\dfrac{2x+\frac{4x^3}{2\sqrt{x^4+1}}}{x^2+\sqrt{x^4+1}}$

So, we'll need the chain rule:

We need the chain rule again:

Use the chain rule.

$\dfrac{\tan x}{2\sqrt{-\log(\cos x)}}$

This requires us to apply the chain rule twice.

Remark: it looks strange to see a negative sign in the argument of a square root. Since the cosine function always gives values that are at most 1, $\log(\cos x)$ is always negative or zero over its domain. So, $\sqrt{\log(\cos x)}$ is only defined for the points where $\cos x=1$ (and so $\log(\cos x) = 0$–this isn't a very interesting function! In contrast, $-\log(\cos x)$ is always positive or zero over its domain – and therefore we can always take its square root.

Use the chain rule to differentiate.

$\dfrac{\sqrt{x^2+4}+x}{x\sqrt{x^2+4}+x^2+4}= \dfrac{1}{\sqrt{x^2+4}}$

Under the chain rule, $\diff{}{x}\log f(x)=\frac{1}{f(x)}f'(x)$. So

You can differentiate this by using the chain rule several times.

$g'(x)=\dfrac{2xe^{x^2}\sqrt{1+x^4}+2x^3}{e^{x^2}\sqrt{1+x^4}+1+x^4}$

Using the chain rule,

Using logarithm rules before you differentiate will make this easier.

$\dfrac{4}{3}$

Using logarithm rules makes this an easier problem:

Using logarithm rules before you differentiate will make this easier.

$f'(x)=\dfrac{3x}{x^2+5}-\dfrac{2x^3}{x^4+10}$

We begin by simplifying:

Now, we differentiate using the chain rule:

Remark: it is a common mistake to write $\log(x^2+4)$ as $\log(x^2)+\log(4)$. These expressions are not equivalent!

First, differentiate using the chain rule and any other necessary rules. Then, plug in $x=2$.

$\dfrac{40}{3}$

We use the chain rule twice, followed by the product rule:

In particular, when $x=2$:

In the text, you are given the derivative $\ds\diff{}{x} a^x$, where $a$ is a constant.

$g'(x)=\pi^x\log \pi+\pi x^{\pi -1}$

In the text, we saw that $\ds\diff{}{x}\left\{a^x\right\}=a^x\log a$ for any constant $a$. So, $\ds\diff{}{x}\left\{\pi^x\right\}=\pi^x\log \pi$.

By the power rule, $\ds\diff{}{x}\left\{x^{\pi}\right\}=\pi x^{\pi-1}$.

Therefore, $g'(x)=\pi^x\log \pi+\pi x^{\pi-1}$.

Remark: we had to use two different rules for the two different terms in $g(x)$. Although the functions $\pi^x$ and $x^\pi$ look superficially the same, they behave differently, as do their derivatives. A function of the form $(\text{constant})^{x}$ is an exponential function and not eligible for the power rule, while a function of the form $x^{\text{constant}}$ is exactly the class of function the power rule applies to.

You'll need to use logarithmic differentiation. Set $g(x)=\log(f(x))$, and find $g'(x)$. Then, use that to find $f'(x)$. This is the method used in the text to find $\ds\diff{}{x} a^x$.

$f'(x) = x^x(\log x + 1)$

We have the power rule to tell us the derivative of functions of the form $x^n$, where $n$ is a constant. However, here our exponent is not a constant. Similarly, in this section we learned the derivative of functions of the form $a^x$, where $a$ is a constant, but again, our base is not a constant! Although the result $\ds\diff{}{x} a^x=a^x\log a$ is not what we need, the method used to differentiate $a^x$ will tell us the derivative of $x^x$.

We'll set $g(x)=\log(x^x)$, because now we can use logarithm rules to simplify:

Now, we can use the product rule to differentiate the right side, and the chain rule to differentiate $\log(f(x))$:

Finally, we solve for $f'(x)$:

Use Question 19 and the base-change formula, $\log_b(a)=\dfrac{\log a}{\log b}$.

$x^x(\log x+1)+\dfrac{1}{x\log 10}$

In Question 19, we saw $\ds\diff{}{x}\left\{x^x\right\}=x^x(\log x+1)$. Using the base-change formula, $\log_{10}(x)=\dfrac{\log x}{\log 10}$. Since $\log_{10}$ is a constant,

To make this easier, use logarithmic differentiation. Set $g(x)=\log(f(x))$, and find $g'(x)$. Then, use that to find $f'(x)$. This is the method used in the text to find $\ds\diff{}{x} a^x$, and again in Question 19.

$f'(x)=\dfrac{1}{4}\left( {\sqrt[4]{\dfrac{(x^4+12)(x^4-x^2+2)}{x^3}}}\right)\left(\dfrac{4x^3}{x^4+12}+\dfrac{4x^3-2x}{x^4-x^2+2}-\dfrac{3}{x}\right)$

Rather than set in with a terrible chain rule problem, we'll use logarithmic differentiation. Instead of differentiating $f(x)$, we differentiate a new function $\log(f(x))$, after simplifying.

Now that we've simplified, we can efficiently differentiate both sides. It is important to remember that we aren't differentiating $f(x)$ directly–we're differentiating $\log(f(x))$.

Our final step is to solve for $f'(x)$:

It was possible to differentiate this function without logarithms, but the logarithms make it more efficient.

To make this easier, use logarithmic differentiation. Set $g(x)=\log(f(x))$, and find $g'(x)$. Then, use that to find $f'(x)$. This is the method used in the text to find $\ds\diff{}{x} a^x$, and again in Question 19.

$f'(x)=(x+1)(x^2+1)^2(x^3+1)^3(x^4+1)^4(x^5+1)^5 \left[\frac{1}{x+1}+\frac{4x}{x^2+1} +\frac{9x^2}{x^3+1}+\frac{16x^3}{x^4+1}+\frac{25x^4}{x^5+1}\right]$

It's possible to do this using the product rule a number of times, but it's easier to use logarithmic differentiation. Set

Now we can use logarithm rules to change $g(x)$ into a form that is friendlier to differentiate:

Now, we differentiate $g(x)$ using the chain rule:

Finally, we solve for $f'(x)$:

It's not going to come out nicely, but there's a better way than blindly applying quotient and product rules, or expanding giant polynomials.

$\left(\dfrac{x^2+2x+3}{3x^4+4x^3+5}\right)\left(\dfrac{1}{x^2+2x+3}-\dfrac{6x^2}{3x^4+4x^3+5}-\dfrac{1}{2(x+1)^2}\right)$

We could do this with quotient and product rules, but it would be pretty painful. Insteady, let's use a logarithm.

Now we have a function that we can differentiate more cleanly than our original function.

Finally, we solve for $f(x)$:

You'll need to use logarithmic differentiation. Set $g(x)=\log(f(x))$, and find $g'(x)$. Then, use that to find $f'(x)$. This is the method used in the text to find $\ds\diff{}{x} a^x$, and again in Question (19).

$f'(x)=(\cos x)^{\sin x}\left[(\cos x) \log (\cos x) - \sin x \tan x\right]$

Since $f(x)$ has the form of a function raised to a functional power, we will use logarithmic differentiation.

Logarithm rules allowed us to simplify. Now, we differentiate both sides of this equation: