Differentiation, Limits, and Critical Point Trouble

[yoursource]

I have 4 problems I'm confused on. They are only practice problems, but if I don't figure these out I'm sure not going to understand the assignment. Not even sure where to start. I do better having practice problems I can see the whole process to before I attempt the assignment. So here they are.

1.) A logistic function is given by N(t)=4900/1+3800e^(-0.05t)
a.) For what value of 't' does this function reach 1/3 of its limiting value?
b.) What is the rate of growth when t equals the answer you found in part (a)? Do not simplify.

2.) Differentiate these functions.
a.) (x^2-6)(lnx)
b.) (sqrt(x))+1/x^2
c.) (3^x)(cubedrt(x+5)-(lnx/sqrt(x))
d.) e^(x^2+2x)/ln(x^4-6)
(for c and d I know I need to do it in steps... but I don't know how.) :|

3.) Assume f(x)=a(x^2-x)(3^(3bx)) has a critical point at x=2 and f(ln3)=2e. Find a and b.
(ahhhh, what? no clue)

4.) Find the equation of the tangent line to the graph of x^4+x^3 when x=5.

Thank you to whoever can walk me through the steps to these problems to get the answers, then I'll better get my assignment!! And be closer to graduation!!

 

[galactus]

Here's a few of them to help out.

1.) A logistic function is given by \(\displaystyle N(t)=\frac{4900}{1+3800e^{-0.05t}}\)
a.) For what value of 't' does this function reach 1/3 of its limiting value?

The limiting value is the value where it levels off. So, \(\displaystyle \lim_{t\to{\infty}}\frac{4900}{1+3800e^{-.05t}}=4900\)

So, we want the value of t that results in 4900/3. \(\displaystyle \frac{4900}{3}=\frac{4900}{1+3800e^{-.05t}}\). Solve for t. Can you?.

2.) Differentiate these functions.
a.) \(\displaystyle (x^{2}-6)(lnx)\)

Product rule: \(\displaystyle (x^{2}-6)(\frac{1}{x})+ln(x)(2x)\). You can simplify a little if you wish.

3.) Assume \(\displaystyle f(x)=a(x^{2}-x)(3^{3bx})\) has a critical point at x=2 and f(ln3)=2e. Find a and b.
(ahhhh, what? no clue)

You find the derivative of the given function, plug in x=2 and set it equal to 0.
Also, plug in ln(3) into your function and set equal to 2e
You have two equations in two unknowns. That should allow you to find a and b.
If yu need help with finding f'(x), let me know. But gives it a go.

4.) Find the equation of the tangent line to the graph of \(\displaystyle x^{4}+x^{3}\) when x=5.

\(\displaystyle f(x)=x^{4}+x^{3}, \;\ f'(x)=4x^{3}+3x^{2}\)

Of course, f'(x) is the slope of your tangent line at some point. We want the slope at x=5. Plug that into f'(x) to find the slope at x=5

Then use y=mx+b. You have m,x,and y. Solve for b and you're done. Oh, you can find y by plugging x=5 into f(x).

 

[yoursource]

Here's a few of them to help out. Sorry I'm still confused. It's in blue where I don't understand.

1.) A logistic function is given by \(\displaystyle N(t)=\frac{4900}{1+3800e^{-0.05t}}\)
a.) For what value of 't' does this function reach 1/3 of its limiting value?

The limiting value is the value where it levels off. So, \(\displaystyle \lim_{t\to{\infty}}\frac{4900}{1+3800e^{-.05t}}=4900\)

So, we want the value of t that results in 4900/3. \(\displaystyle \frac{4900}{3}=\frac{4900}{1+3800e^{-.05t}}\). Solve for t. Can you?. I don't know how to solve for t when it is in a fraction and superscript like this.

[quote:v90041ea]2.) Differentiate these functions.
a.) \(\displaystyle (x^{2}-6)(lnx)\)

Product rule: \(\displaystyle (x^{2}-6)(\frac{1}{x})+ln(x)(2x)\). You can simplify a little if you wish. Is this all there is to it then? Is there more to the process before this answer?

3.) Assume \(\displaystyle f(x)=a(x^{2}-x)(3^{3bx})\) has a critical point at x=2 and f(ln3)=2e. Find a and b.
(ahhhh, what? no clue)

You find the derivative of the given function, plug in x=2 and set it equal to 0.
Also, plug in ln(3) into your function and set equal to 2e
You have two equations in two unknowns. That should allow you to find a and b.
If yu need help with finding f'(x), let me know. But gives it a go.

I don't know how to find the derivative of a problem like this where there is a subscript and the a out front... does that just go away?

4.) Find the equation of the tangent line to the graph of \(\displaystyle x^{4}+x^{3}\) when x=5.

\(\displaystyle f(x)=x^{4}+x^{3}, \;\ f'(x)=4x^{3}+3x^{2}\)

Of course, f'(x) is the slope of your tangent line at some point. We want the slope at x=5. Plug that into f'(x) to find the slope at x=5

Then use y=mx+b. You have m,x,and y. Solve for b and you're done. Oh, you can find y by plugging x=5 into f(x).[/quote:v90041ea]

Thanks for all your help!!

 

[galactus]

#1: To begin with, turn it upside and down and write as: \(\displaystyle \frac{3}{4900}=\frac{1+3800e^{-.05t}}{4900}\)

\(\displaystyle 3=1+3800e^{-.05t}\)

\(\displaystyle \frac{1}{1900}=e^{-.05t}\)

\(\displaystyle ln(\frac{1}{1900})=-.05t\)

\(\displaystyle t=\frac{ln(\frac{1}{1900})}{-.05}\approx{151}\)

See there?.Not that bad. Just some algebra.

#2: some of them may involve the chain rule as well. This one was straightforward.

#3. if you're a rookie at differentiation, this one will probably be tricky. The a is a constant. Treat it as such.

Differentiating and factoring, we get: \(\displaystyle a27^{bx}(3bln(3)x^{2}-3bxln(3)+2x-1)\)

I would practice some logarithmic differentiation and the product rule, chain rule, etc.

I can do only so much here.

 

[stapel]

So, we want the value of t that results in 4900/3. \(\displaystyle \frac{4900}{3}=\frac{4900}{1+3800e^{-.05t}}\). Solve for t. Can you?.

I don't know how to solve for t when it is in a fraction and superscript like this.

Ouch! They were supposed to have taught you algebra before they put you in a calculus class! :shock:

Okay, so you're not familiar with how to solve exponential equations. Are you familiar with exponentials and logarithms at all, or should we start the lessons there? (You can solve linear, quadratic, and rational equations, though, right?)

2.) Differentiate these functions.
a.) \(\displaystyle (x^{2}-6)(lnx)\)

Product rule: \(\displaystyle (x^{2}-6)(\frac{1}{x})+ln(x)(2x)\). You can simplify a little if you wish.

Is this all there is to it then? Is there more to the process before this answer?

Erm... What do you mean by "more to" the final act of simplifying...? Or are you not familiar with the Product Rule, and need lessons on what this means, so you can understand how the tutor did the exercise for you? :?:

3.) Assume \(\displaystyle f(x)=a(x^{2}-x)(3^{3bx})\) has a critical point at x=2 and f(ln3)=2e. Find a and b.

You find the derivative of the given function, plug in x=2 and set it equal to 0. Also, plug in ln(3) into your function and set equal to 2e. You have two equations in two unknowns. That should allow you to find a and b.

I don't know how to find the derivative of a problem like this where there is a subscript and the a out front... does that just go away?

Um... When the tutor explained, step by step, how to work with the "a" and "b" constants, this was sort of a hint that you must, as the exercise specified, find the values of "a" and "b". This would of course be impossible if they somehow magically disappeared. :!:

As posted, there is no "subscript", so I'm afraid we can't help there. :|

So it would seem that you aren't familiar with constants or variables, or what a system of equations might be. These topics are covered in algebra, but not usually in the same semester, so you're looking at possibly weeks of material. It might be time to start thinking about hiring a qualified local tutor, so you can work face-to-face with someone who can help you get caught up on the all the missing courses' worth of content. :idea:

Please reply with clarification of the various topics on which you are needing instruction. As it would appear that they have skipped over a year or so of material, please be complete. Thank you!

Eliz.

 

[yoursource]

So, we want the value of t that results in 4900/3. \(\displaystyle \frac{4900}{3}=\frac{4900}{1+3800e^{-.05t}}\). Solve for t. Can you?.

I don't know how to solve for t when it is in a fraction and superscript like this.

Ouch! They were supposed to have taught you algebra before they put you in a calculus class! :shock:

Okay, so you're not familiar with how to solve exponential equations. Are you familiar with exponentials and logarithms at all, or should we start the lessons there? (You can solve linear, quadratic, and rational equations, though, right?) ( I have been taught all of this, it's just been so long I'm confusing myself more than anything)

2.) Differentiate these functions.
a.) \(\displaystyle (x^{2}-6)(lnx)\)

Product rule: \(\displaystyle (x^{2}-6)(\frac{1}{x})+ln(x)(2x)\). You can simplify a little if you wish.

Is this all there is to it then? Is there more to the process before this answer?

Erm... What do you mean by "more to" the final act of simplifying...? Or are you not familiar with the Product Rule, and need lessons on what this means, so you can understand how the tutor did the exercise for you? :?: (I get this part now)

3.) Assume \(\displaystyle f(x)=a(x^{2}-x)(3^{3bx})\) has a critical point at x=2 and f(ln3)=2e. Find a and b.

You find the derivative of the given function, plug in x=2 and set it equal to 0. Also, plug in ln(3) into your function and set equal to 2e. You have two equations in two unknowns. That should allow you to find a and b.

I don't know how to find the derivative of a problem like this where there is a subscript and the a out front... does that just go away?

Um... When the tutor explained, step by step, how to work with the "a" and "b" constants, this was sort of a hint that you must, as the exercise specified, find the values of "a" and "b". This would of course be impossible if they somehow magically disappeared. :!:

As posted, there is no "subscript", so I'm afraid we can't help there. :| (Sorry I meant superscript, the part with 3^(3bx) I'm just so rusty, it's been since High School, and I'm a senior in College.)

So it would seem that you aren't familiar with constants or variables, or what a system of equations might be. These topics are covered in algebra, but not usually in the same semester, so you're looking at possibly weeks of material. It might be time to start thinking about hiring a qualified local tutor, so you can work face-to-face with someone who can help you get caught up on the all the missing courses' worth of content. :idea:

Please reply with clarification of the various topics on which you are needing instruction. As it would appear that they have skipped over a year or so of material, please be complete. Thank you!

Eliz.

The parts I still need help on are 2. (b, c, d) and then 4, the derivative looks so confusing.

 

[galactus]

2d:

\(\displaystyle f(x)=\frac{e^{x^{2}+2x}}{ln(x^{4}-6)}\)

You need the quotient rule and chain rule.

\(\displaystyle \frac{ln(x^{4}-6)\overbrace{\overbrace{(2x+2)e^{x^{2}+2x}}^{\text{chain rule}}}^{\text{derivative of\\the e term}}-e^{x^{2}+2x}\frac{(4x^{3}}{x^{4}-6})}{(ln(x^{4}-6))^{2}}\)

Now, tidy it up by factoring out a common term and what not.

As for #4, that is about the easiest problem here. Just plug and chug.

 

[yoursource]

Thank you on 2d... and sorry, I didn't mean 4, you're right that one was easy. I meant 3 still. I know you showed me it after it is differentiated and factored, but I'd like to see the steps to get to that if it's possible. I think I'd understand it better that way.

 

[skeeter]

2.) Differentiate these functions.

b.) (sqrt(x))+1/x^2 this is a straight-forward derivative ... what did you get?

c.) (3^x)(cubedrt(x+5)-(lnx/sqrt(x))

you'll have to be careful with this one ... product rule and quotient rule are involved. remember that
\(\displaystyle \frac{d}{dx}[3^x] = 3^x \cdot \ln{3}\)

d.) e^(x^2+2x)/ln(x^4-6)

quotient rule coupled with the chain rule ...
\(\displaystyle \frac{d}{dx} \left(\frac{e^u}{\ln{v}}\right) = \frac{\ln{v} \cdot e^u \frac{du}{dx} - e^u \frac{\frac{dv}{dx}}{v}}{(\ln{v})^2}\)


4.) Find the equation of the tangent line to the graph of x^4+x^3 when x=5.

you need two items ... the point on the curve (5,f(x)) and the slope at that point on the curve, f'(5). then use the point-slope form of a linear equation to find the tangent line equation.

 

[galactus]

Well, for the third one, you have to remember that \(\displaystyle \frac{d}{dx}[b^{u}]=b^{u}ln(b)\cdot\frac{du}{dx}\), where u is a differentiable function of x. Which is what you have. Follow the formula for that portion and use the product rule. In your case, \(\displaystyle u=3bx\)
Remember, you can write \(\displaystyle 3^{3bx}=27^{bx}\) if need be.

 

[yoursource]

b.) I got: (1/2x^(5/2)) - (2/x^3)(sqrt(x)+1) am i way off?

c.) I don't even know how to begin to do this with the product and quotient rule. :?:

 

[skeeter]

b.) I got: (1/2x^(5/2)) - (2/x^3)(sqrt(x)+1) am i way off?

first of all, your original problem should have been written using proper notation.

as you wrote it ... (sqrt(x))+1/x^2 means
\(\displaystyle \sqrt{x} + \frac{1}{x^2}\)

when you actually meant \(\displaystyle \frac{\sqrt{x} + 1}{x^2}\)

quotient rule ...

\(\displaystyle \frac{x^2 \frac{1}{2\sqrt{x}} - 2x(\sqrt{x} + 1)}{x^4}\)



c.) I don't even know how to begin to do this with the product and quotient rule. :?:

(3^x)(cubedrt(x+5)-(lnx/sqrt(x)) ... I'll set this one up for you ...
let u = 3[sup:1w5s84eb]x[/sup:1w5s84eb], v = cbrt(x+5), y = lnx, and z = sqrt(x) ...

\(\displaystyle \frac{d}{dx} \left[u \left(v - \frac{y}{z} \right) \right] = u \left(v' - \frac{zy' - yz'}{z^2} \right) + u' \left(v - \frac{y}{z} \right)\)

I'll let you slug your way thru the algebra.