MHB Logarithmic Differentiation And Exponential Growth Questions

[ardentmed]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For the first one (part a), I went through the steps for logarithmic differentiation by using the ln laws to separate terms, and then taking the derivative of both sidews. After isolating dy/dx, I computed:

dy/dx = [3^(-5x) (2x-3)^4/3]/[(7x-5)^2/5 * (3x-7)^3/4] * (-5xln(3)/3 + 8/(3(2x-3)) - (14/5(7x-5)) - (9/4(3x-7))

As for 1b, I did the same thing, bringing down the exponent this time and paying careful attention to the cscx when differentiating. Ultimately, I got:

dy/dx = 3csc(3x)^(3/x) * [(-1/x^2)ln(csc(3x)) - 3cot(3x))

And finally, for the "yam" question. I used exponential growth (albeit the altered form for temperature change). So:

dT/dt = (250-ae^(-bt)) dT/dt

Ultimately, I got b=-.03 and a = 100e^(9/100)

I verified the formula:
250-100e^(9/100 - 3t/100) = T(t)
with t=30 and got 150 degrees (which I'm guessing means it is correct, right? I'm not too sure though.)Thanks in advance.

 

[MarkFL]

I recommend you take some time to learn how to use $\LaTeX$ to make your math easier to read, and also to post your work rather than just your answers, because if your answer is wrong, we have no idea why. Also, I find some of the images to be hard to read. You want to make it as easy as possible for people to provide help. If they are struggling to read the questions, and struggling to make sense of results consisting of long lines of plain text, then people are going to be less likely to help.

So, that being said, let's take a look at the first question, where we are to use logarithmic differentiation:

$$y=\frac{3^{-5x}(2x-3)^{\frac{4}{3}}}{(7x-5)^{\frac{2}{5}}(3x-7)^{\frac{3}{4}}}$$

Taking the natural log of both sides and applying the logarithmic rules, we may write:

$$\ln(y)=-5x\ln(3)+\frac{4}{3}\ln(2x-3)-\frac{2}{5}\ln(7x-5)-\frac{3}{4}\ln(3x-7)$$

Now, differentiating with respect to $x$, we obtain:

$$\frac{1}{y}\cdot\d{y}{x}=-5\ln(3)+\frac{8}{3(2x-4)}-\frac{14}{5(7x-5)}-\frac{9}{4(3x-7)}$$

Do you see where you made an error differentiating the first term on the right? You just have the constant $$-5\ln(3)$$ times $x$, and so you apply the rule:

$$\frac{d}{dx}(kx)=k$$

to that term.

For 1b), your result is close to, but not quite the same as that returned by W|A, so let's walk through it...

We are given:

$$y=\left(\csc(3x)\right)^{\frac{3}{x}}$$

Taking the natural log of both sides, and applying the rules of logs, we obtain:

$$\ln(y)=\frac{3}{x}\ln\left(\csc(3x)\right)$$

Differentiating with respect to $x$, we find:

$$\frac{1}{y}\cdot\d{y}{x}=\frac{3}{x}\cdot\frac{-3\cot(3x)\csc(3x)}{\csc(3x)}-\frac{3}{x^2}\ln\left(\csc(3x)\right)$$

Simplify a bit and factor:

$$\frac{1}{y}\cdot\d{y}{x}=-\frac{3}{x^2}\left(3x\cot(3x)+\ln\left(\csc(3x)\right)\right)$$

I will leave you to finish...:D

 

[ardentmed]

I recommend you take some time to learn how to use $\LaTeX$ to make your math easier to read, and also to post your work rather than just your answers, because if your answer is wrong, we have no idea why. Also, I find some of the images to be hard to read. You want to make it as easy as possible for people to provide help. If they are struggling to read the questions, and struggling to make sense of results consisting of long lines of plain text, then people are going to be less likely to help.

So, that being said, let's take a look at the first question, where we are to use logarithmic differentiation:

$$y=\frac{3^{-5x}(2x-3)^{\frac{4}{3}}}{(7x-5)^{\frac{2}{5}}(3x-7)^{\frac{3}{4}}}$$

Taking the natural log of both sides and applying the logarithmic rules, we may write:

$$\ln(y)=-5x\ln(3)+\frac{4}{3}\ln(2x-3)-\frac{2}{5}\ln(7x-5)-\frac{3}{4}\ln(3x-7)$$

Now, differentiating with respect to $x$, we obtain:

$$\frac{1}{y}\cdot\d{y}{x}=-5\ln(3)+\frac{8}{3(2x-4)}-\frac{14}{5(7x-5)}-\frac{9}{4(3x-7)}$$

Do you see where you made an error differentiating the first term on the right? You just have the constant $$-5\ln(3)$$ times $x$, and so you apply the rule:

$$\frac{d}{dx}(kx)=k$$

to that term.

For 1b), your result is close to, but not quite the same as that returned by W|A, so let's walk through it...

We are given:

$$y=\left(\csc(3x)\right)^{\frac{3}{x}}$$

Taking the natural log of both sides, and applying the rules of logs, we obtain:

$$\ln(y)=\frac{3}{x}\ln\left(\csc(3x)\right)$$

Differentiating with respect to $x$, we find:

$$\frac{1}{y}\cdot\d{y}{x}=\frac{3}{x}\cdot\frac{-3\cot(3x)\csc(3x)}{\csc(3x)}-\frac{3}{x^2}\ln\left(\csc(3x)\right)$$

Simplify a bit and factor:

$$\frac{1}{y}\cdot\d{y}{x}=-\frac{3}{x^2}\left(3x\cot(3x)+\ln\left(\csc(3x)\right)\right)$$

I will leave you to finish...:D

Alright, thanks a ton!

I got $$\cdot\d{y}{x}= f(x)[-5\ln(3)+\frac{8}{3(2x-3)}-\frac{14}{5(7x-5)}-\frac{9}{4(3x-7)}]$$

Where f(x) is just the original function, y= 3^ ..

Should I simplify it further by taking the lowest common denominator and multiplying it out, or is that enough?
Thanks again.

 

[MarkFL]

If it were me, I would try to simplify as much as possible, but that's really a matter of taste and preference. I like results to be as simple as possible, but sometimes this can be a lot of work and you aren't guaranteed it will pay off either. It really depends on how much time you have. :D