I let the mistake of forgetting to put parentheses when appropriate pass under the radar for this homework. But you might not be so lucky in future homeworks.
3.6 #6
$f(x)=\log_5(xe^x)=\frac{\ln(xe^x)}{\ln(5)}=\frac{\ln(x)+\ln(e^x)}{\ln(5)}=\frac{\ln(x)+x}{\ln(5)}$
That was just algebraic manipulation to make the problem easier.
$\frac{df}{dx}(x)=\frac{1}{\ln(5)}\left(\frac{1}{x}+1\right)$
An acceptable final answer is:
$\frac{df}{dx}(x)=\frac{1}{\ln(5)}\left(\frac{x+1}{x}\right)$
An unacceptable final answer is:
$\frac{df}{dx}(x)=\frac{1}{\ln(5)}\left(\frac{xe^x+e^x}{xe^x}\right)$
3.6 #11
$f(x)=\ln\left(\frac{(2t+1)^3}{(3t-1)^4}\right)=3\ln(2t+1)-4\ln(3t-1)$
That was just algebraic manipulation to make the problem easier.
$\frac{df}{dx}(x)=\frac{3}{2t+1}\cdot 2-\frac{4}{3t-1}\cdot 3$
$\frac{df}{dx}(x)=\frac{6}{2t+1}-\frac{12}{3t-1}$
Error:
$3\ln(2t+1)-4\ln(3t-1)\neq \frac{3}{2t+1}\cdot 2-\frac{4}{3t-1}\cdot 3$
Correction: $\frac{d}{dx}\left[3\ln(2t+1)-4\ln(3t-1)\right] = \frac{3}{2t+1}\cdot 2-\frac{4}{3t-1}\cdot 3$
Point Loss:
Using chain rule and quotient rule. Using this method to find the derivative requires more computation than the solution above. It will more likely result in an algebraic error. In addition, it shows the student for some reason does not know how the easier method, which doesn't bode well.
Note:
Changing $f(x)$ into $\ln\left[(2t+1)^3\right]-\ln\left[(3t-1)^4\right]$ and then differentiating is a better alternative over using "the chain rule and then the quotient rule." However, taking the extra step to bring down the 3 and 4 will save work and avoid errors.
3.6 #12
$h(x)=\ln(x+\sqrt{x^2-1})$
$\frac{dh}{dx}(x)=\frac{1}{x+\sqrt{x^2-1}}\left(1+\frac{1}{2}\frac{1}{\sqrt{x^2-1}}(2x)\right)$
Note the difference between needing to simplify the unacceptable answer in 3.6 #6 and not needing to simplify the expression here. Basically, simplifying should be done when it offers clarity or simplicity. Both the answer and listed acceptable answer in 3.6 #6 lack the extra terms $e^x$ which can clearly be canceled.
Types of errors:
$\ln(x+\sqrt{x^2-1})\neq\ln(x)+\ln(\sqrt{x^2-1})$
3.6 #16
$y=\frac{1}{\ln(x)}=[\ln(x)]^{-1}$
$\frac{dy}{dx}=-[\ln(x)]^{-2}\frac{1}{x}=-\frac{1}{x[\ln(x)]^2}$
Note: When I take the derivative of $y=\frac{1}{f(x)}$ I use the power rule and chain rule to get $\frac{dy}{dx}=-\frac{1}{[f(x)]^2}\frac{df}{dx}$
Note: Sometimes I write $\frac{dy}{dx}(x)$ to remind myself that the derivative is a function of $x$. And sometimes I don't, either because I forget or because I think it might be confusing. In general, we have,
for $y=y(x)$,
$\frac{dy}{dx}(x)=\frac{dy}{dx}=y^{\prime}(x)=y^{\prime}$.
If you go on to take multivariable calculus, you would have the following corresponding notation:
for $f=f(x_1,x_2,\dots,x_n)$,
$\frac{\partial f}{\partial x_i}(x_1,\dots,x_n)=\frac{df}{dx}=f_{x_1}(x_1,\dots,x_n)=f_{x_1}$
3.6 #20
$\frac{dy}{dx}=2[\ln(1+e^x)]\frac{1}{1+e^x}e^x=\frac{2 e^x \ln(1+e^x)}{1+e^x}$
Point Loss:
Not simplifying $\frac{d}{dx}e^x$
Correction: $\frac{d}{dx}e^x=e^x$
Error:
$[\ln(1+e^x)]^2\neq 2\ln(1+e^x)$
There is a difference between $[\ln(1+e^x)]^2$ and $\ln\left[(1+e^x)^2\right]$
For the second we have $\ln\left[(1+e^x)^2\right]=2\ln(1+e^x)$
$\frac{d}{dx}(\ln(1+e^x))^2\neq 2(\ln(1+e^x))^{\prime}(1+e^x)^{\prime}$
There is a difference between $(\ln(1+e^x))^{\prime}$ and $(\ln(x))^\prime|_{1+e^x}$
The first means take the derivative of $\ln(1+e^x)$ and the second means take the derivative of $\ln(x)$ and then plug in $1+e^x$.
Using other notation the above error would be:
$\frac{d}{dx}(\ln(1+e^x))^2\neq 2\frac{d}{dx}(\ln(1+e^x))\frac{d}{dx}(1+e^x)$
There is a difference between $\left(\frac{d}{dx}(\ln(1+e^x))\right)(x)$ and $\left(\frac{d}{dx}(\ln(x))\right)(1+e^x)$.
Note that here I write $\left(\frac{d}{dx}(\ln(1+e^x))\right)(x)$ when we would usually write $\frac{d}{dx}(\ln(1+e^x))$.
In any case, we usually do multiple steps in the chain rule in our head, but be careful not to confuse the steps.
3.6 #24
Using the quotient rule:
$\frac{dy}{dx}=\frac{\frac{1}{x}x^2-\ln(x)(2x)}{x^4}=\frac{1-2\ln(x)}{x^3}$
$\frac{d^2y}{dx^2}=\frac{\frac{-2}{x}x^3-(1-2\ln(x))(3x^2)}{x^6}=\frac{-2-(3-6\ln(x))}{x^4}=\frac{6\ln(x)-5}{x^4}$
Using the product rule:
$\frac{dy}{dx}=\frac{1}{x}\frac{1}{x^2}+\frac{-2\ln(x)}{x^3}=\frac{1}{x^3}+\frac{-2\ln(x)}{x^3}$
$\frac{d^2y}{dx^2}=\frac{-3}{x^4}+\frac{-2}{x^4}+\frac{6\ln(x)}{x^4}=\frac{6\ln(x)-5}{x^4}$
3.6 #30
$y=\ln(\ln(\ln(x)))$
$\frac{dy}{dx}=\frac{1}{\ln(\ln(x))}\frac{1}{\ln(x)}\frac{1}{x}$
3.6 #40
$y=\sqrt[4]{\frac{x^2+1}{x^2-1}}$
Take the natural log of both sides and use implicit differentiation:
$\ln(y)=\ln\left(\sqrt[4]{\frac{x^2+1}{x^2-1}}\right)$
$=\ln\left(\frac{x^2+1}{x^2-1}\right)^{\frac{1}{4}}$
$=\frac{1}{4}\ln\left(\frac{x^2+1}{x^2-1}\right)$
$=\frac{1}{4}\left(\ln(x^2+1)-\ln(x^2-1)\right)$
The above was just taking natural log of both sides and algebraic manipulation.
Now for the implicit differentiation:
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{4}\left(\frac{1}{x^2+1}(2x)-\frac{1}{x^2-1}(2x)\right)$
$\frac{dy}{dx}=y\left(\frac{1}{4}\left(\frac{1}{x^2+1}(2x)-\frac{1}{x^2-1}(2x)\right)\right)$
$=\left(\sqrt[4]{\frac{x^2+1}{x^2-1}}\right)\left(\frac{1}{4}\left(\frac{1}{x^2+1}(2x)-\frac{1}{x^2-1}(2x)\right)\right)$
$=\left(\sqrt[4]{\frac{x^2+1}{x^2-1}}\right)\left(\frac{1}{2}\left(\frac{x}{x^2+1}-\frac{x}{x^2-1}\right)\right)$
The above is correct, but in fact the second factor can be simplified slightly for a more aesthetically pleasing answer:
$\frac{dy}{dx}=\left(\sqrt[4]{\frac{x^2+1}{x^2-1}}\right)\left(\frac{1}{2}\left(\frac{-x}{x^4-1}\right)\right)=\frac{1}{2}\frac{-x}{x^4-1}\sqrt[4]{\frac{x^2+1}{x^2-1}}$
3.6 #48
$y=[\ln(x)]^{\cos(x)}$
$\ln(y)=\ln\left([\ln(x)]^{\cos(x)}\right)=\cos(x)\ln(\ln(x))$
Differentiate both sides:
$\frac{1}{y}\frac{dy}{dx}=-\sin(x)\ln(\ln(x))+\cos(x)\left[\frac{1}{\ln(x)}\frac{1}{x}\right]$
Then:
$\frac{dy}{dx}=y\left[-\sin(x)\ln(\ln(x))+\cos(x)\frac{1}{x\ln(x)}\right]$
$=[\ln(x)]^{\cos(x)}\left[-\sin(x)\ln(\ln(x))+\frac{\cos(x)}{x\ln(x)}\right]$
3.6 #50
$x^y=y^x$
$y\ln(x)=x\ln(y)$
$\frac{dy}{dx}\ln(x)+y\frac{1}{x}=\ln(y)+x\frac{1}{y}\frac{dy}{dx}$
$\frac{dy}{dx}(\ln(x)-\frac{x}{y})=\ln(y)-\frac{y}{x}$
Then
$\frac{dy}{dx}=\left(\ln(y)-\frac{y}{x}\right)/\left(\ln(x)-\frac{x}{y}\right)$
Alternatively:
$e^{y\ln(x)}=e^{x\ln(y)}$ gets differentiated to obtain
$e^{y\ln(x)}(\frac{dy}{dx}\ln(x)+y\frac{1}{x})=e^{x\ln(y)}(\ln(y)+x\frac{1}{y}\frac{dy}{dx})$
$x^y(\frac{dy}{dx}\ln(x)+\frac{y}{x})=y^x(\ln(y)+\frac{x}{y}\frac{dy}{dx})$
$\frac{dy}{dx}(x^y\ln(x)-y^x\frac{x}{y})=y^x\ln(y)-x^y\frac{y}{x}$
Then
$\frac{dy}{dx}=\left(y^x\ln(y)-x^y\frac{y}{x}\right)/\left(x^y\ln(x)-y^x\frac{x}{y}\right)$
Finally use the fact that $x^y=y^x$ to obtain the answer we got above:
$\frac{dy}{dx}(x^y\ln(x)-y^x\frac{x}{y})=y^x\ln(y)-x^y\frac{y}{x}$
Error:
For a constant $c$ we can do $\frac{d}{dx}x^c=cx^{c-1}$
but $\frac{d}{dx}(x^y)\neq yx^{y-1}$ because $y$ depends on $x$.
Correction: $\frac{d}{dx}(x^y)=\frac{d}{dx}(e^{y\ln(x)}=(e^{y\ln(x)}\left(y^{\prime}\ln(x)+\frac{y}{x}\right)$
For a constant $c$ we can do $\frac{d}{dx}y^c=cy^{c-1}\frac{dy}{dx}$
but $\frac{d}{dx}(y^x)\neq xy^{x-1}y^{\prime}$ because $y$ depends on $x$.
Correction:
$\frac{d}{dx}(y^x)=\frac{d}{dx}(e^{x\ln(y)}=(e^{x\ln(y)}\left(\ln(y)+\frac{x}{y}y^{\prime}\right)$
For a constant $c$ we can do $\frac{d}{dx}(c^y)=c^y\ln(c)\frac{dy}{dx}$.
But $\frac{d}{dx}(x^y)\neq x^y\ln(x)\frac{dy}{dx}$ because $x$ depends on $x$.
Also $\frac{d}{dx}(y^x)\neq y^x\ln(y)\frac{dx}{dx}$ because $y$ depends on $x$.
Correction:
See above corrections.
3.9 #4
3.9 #30
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