Calc - Derivatives and Differentiation of Logs

[chops369]

Homework Statement

I have a few problems that are giving me some trouble:

1. Take the derivative of xe^-4x

2. Find dy/dx and evaluate the slope for the curve e^{y^3} - 2x⁴ + y² = 3 at (8,0)

3. Find dy/dx and evaluate the slope for the curve e^-y - 4 = x² + 1 at (-2,2)

Homework Equations

N/A

The Attempt at a Solution

1. I'm OK with taking these types of derivatives, but the x out in front is throwing me off. I'm not quite sure what to do with it. For now, I have this as my answer: -4xe^-4x

2. I'm not going to type out every step, but I took d/dx of both sides and eventually came up with dy/dx(3ye^{y^3} + 2y) = 8x³

Then I solved for dy/dx to get 8x³ / (3ye^{y^3} + 2y)

The only thing that gives me reason to believe my answer is incorrect is the fact that when I plug in the values for x and y from the coordinate given, I get 4096 / 0, which would mean the slope is undefined. For some reason this doesn't seem right.

3. I took d/dx of both sides and eventually got -e^-ydy/dx = 2x

Then I solved for dy/dx to get 2x / (-e^-y)

Then I plugged in the coordinate values and got a slope of 29.6. Is this correct? I wasn't sure if it was OK to move the terms as such in the beginning: e^-y - x² = 5

 

[zachzach]

Homework Statement

I have a few problems that are giving me some trouble:

1. Take the derivative of xe^-4x

2. Find dy/dx and evaluate the slope for the curve e^{y^3} - 2x⁴ + y² = 3 at (8,0)

3. Find dy/dx and evaluate the slope for the curve e^-y - 4 = x² + 1 at (-2,2)

Homework Equations

The Attempt at a Solution

1. I'm OK with taking these types of derivatives, but the x out in front is throwing me off. I'm not quite sure what to do with it. For now, I have this as my answer: -4xe^-4x

2. I'm not going to type out every step, but I took d/dx of both sides and eventually came up with dy/dx(3ye^{y^3} + 2y) = 8x³

Then I solved for dy/dx to get 8x³ / (3ye^{y^3} + 2y)

The only thing that gives me reason to believe my answer is incorrect is the fact that when I plug in the values for x and y from the coordinate given, I get 4096 / 0, which would mean the slope is undefined. For some reason this doesn't seem right.

3. I took d/dx of both sides and eventually got -e^-ydy/dx = 2x

Then I solved for dy/dx to get 2x / (-e^-y)

Then I plugged in the coordinate values and got a slope of 29.6. Is this correct? I wasn't sure if it was OK to move the terms as such in the beginning: e^-y - x² = 5

1. You must use the product rule With f = x and g = e^-4x.

2. Are you using implicit differentiation? If not you should.

3. You can move the terms, you are just simplifying the equation. Personally I would isolate the y variable, take the ln of both sides and then differentiate.

 

[Mark44]

Homework Statement

I have a few problems that are giving me some trouble:

1. Take the derivative of xe^-4x

2. Find dy/dx and evaluate the slope for the curve e^{y^3} - 2x⁴ + y² = 3 at (8,0)

3. Find dy/dx and evaluate the slope for the curve e^-y - 4 = x² + 1 at (-2,2)

Homework Equations

N/A

The Attempt at a Solution

1. I'm OK with taking these types of derivatives, but the x out in front is throwing me off. I'm not quite sure what to do with it. For now, I have this as my answer: -4xe^-4x

Nope. You need to use the product rule.

2. I'm not going to type out every step, but I took d/dx of both sides and eventually came up with dy/dx(3ye^{y^3} + 2y) = 8x³

It doesn't look like you used the chain rule correctly. E.g., d/dx(e^y2) = e^y2 * d/dx(y²) = e^y2 * 2y*y'.

Then I solved for dy/dx to get 8x³ / (3ye^{y^3} + 2y)

The only thing that gives me reason to believe my answer is incorrect is the fact that when I plug in the values for x and y from the coordinate given, I get 4096 / 0, which would mean the slope is undefined. For some reason this doesn't seem right.

3. I took d/dx of both sides and eventually got -e^-ydy/dx = 2x

Then I solved for dy/dx to get 2x / (-e^-y)

Then I plugged in the coordinate values and got a slope of 29.6. Is this correct? I wasn't sure if it was OK to move the terms as such in the beginning: e^-y - x² = 5

This one looks OK. Sure you can move terms around from one side to the other of an equation and then take the derivative.

 

[chops369]

Nope. You need to use the product rule.

Really? Why wouldn't the x just go away? For example, d/dx (ex) = e ?

 

[Mark44]

Really? Why wouldn't the x just go away? For example, d/dx (ex) = e ?

d/dx(k*f(x)) = k *d/dx(f(x)), so d/dx(ex) = e*d/dx(x) = e

But the problem I referred to had xe^-4x, so the product rule has to be used. Although e is a constant, e^-4x is a function of x.

 

[chops369]

d/dx(k*f(x)) = k *d/dx(f(x)), so d/dx(ex) = e*d/dx(x) = e

But the problem I referred to had xe^-4x, so the product rule has to be used. Although e is a constant, e^-4x is a function of x.

Oh, OK I see.

I'm not sure if I used the product rule correctly in this case. I'm still a bit shaky with some of these derivatives.

Here is what I did:

f '(x) = 1 * e^-4x + x * -4e^-4x
Then simplifying to get e^-4x - 4xe^-4x as my final answer.

 

[Mark44]

That's more like it.

 

[chops369]

OK, great. I'm still having major trouble with the second problem. This e^y3 is driving me insane.

Did you say that I need to use the chain rule on it? I'm really confused about this.

I'm don't really understand the example of the chain rule you posted earlier. To me, it looks like it suggests that d/dx (e^y3) = e^y3 * 2y * 2y ? Also, for this type of problem, how would you account for adding dy/dx variables in, as in implicit differentiation? Would you just proceed as normal?

 

[zachzach]

OK, great. I'm still having major trouble with the second problem. This e^y3 is driving me insane.

Did you say that I need to use the chain rule on it? I'm really confused about this.

dy/dx[e^y3] = ?

dy/dx [e^y] = e^y * dy/dx Does this help you?

 

[zachzach]

I'm don't really understand the example of the chain rule you posted earlier. To me, it looks like it suggests that d/dx (e^y3) = e^y3 * 2y * 2y ? Also, for this type of problem, how would you account for adding dy/dx variables in, as in implicit differentiation? Would you just proceed as normal?

d/dx (e^y3) = e^y3 * 2y * 2y is not correct.

You know that d/du[e^u] = e^u*u'. So what is the derivative of y³? Namely what is d/dx [y³] =? Then you can just multiply that by e^{y3 and that is your answer. There will be a dy/dx in your answer.}

 

[chops369]

d/dx (e^y3) = e^y3 * 2y * 2y is not correct.

You know that d/du[e^u] = e^u*u'. So what is the derivative of y³? Namely what is d/dx [y³] =? Then you can just multiply that by e^{y3 and that is your answer. There will be a dy/dx in your answer.}

^{Is it (3y2 * dy/dx) * ey3, then? But isn't this not the chain rule?}

 

[wisvuze]

y is an implicit function of x, so when you differentiate d/dx e^(y^3) you have to use the chain rule for e ^ ( y^3 ) [the ouside e^__ ] and then the chain rule again for y^3

 

[zachzach]

Is it (3y² * dy/dx) * e^{y3, then? But isn't this not the chain rule?}

^{Yes that is the correct answer. It is the chain rule if you let u = ey3 then by the chain rule du/dx = du/dy * dy/dx.
du/dy = ey3*(3y2) and dy/dx = dy/dx.}

 

[chops369]

Ok, now I'm stuck again, but only on something minor.

I now have 3y²(dy/dx) * e^y3 + 2y(dy/dx) = 8x³

If I want to factor out a dy/dx, how can I combine the 3y² with the e^y3 ? Do I just lump them together, i.e. 3y²e^y3 ?

If so, my original answer was correct for the most part and I am still left with a slope of 4096 / 0 = und.

 

[zachzach]

Ok, now I'm stuck again, but only on something minor.

I now have 3y²(dy/dx) * e^y3 + 2y(dy/dx) = 8x³

If I want to factor out a dy/dx, how can I combine the 3y² with the e^y3 ? Do I just lump them together, i.e. 3y²e^y3 ?

3y²(dy/dx) * e^y3 + 2y(dy/dx) = 8x³

Like you would any other variable:

dy/dx(3y²*e^y3+2y)

So yes, the the order of multiplication does not matter and dy/dx is multiplying that argument.

3y²(dy/dx) * e^y3 = 3y²*(dy/dx) * e^y3

=3y²*e^y3*(dy/dx)