Derivative of y = cos(a^3 + x^3)

[illjazz]

Homework Statement

Find the derivative of the function y = cos(a^3 + x^3).

Homework Equations

- Chain rule
- Power rule

The Attempt at a Solution

This is driving me insane.. so here's what I have.

y = cos(a^3 + x^3)

y' = -sin(a^3 + x^3) * d/dx (a^3 + x^3)

y' = -sin(a^3 + x^3) * (3a^2 + 3x^2

y' = -(3a^2)(sin(a^3 + x^3)) - (3x^2)(sin(a^3 + x^3))

I would like to think that that's where this ends.. but my book claims the result is

y' = -(3x^2)(sin(a^3 + x^3))

How? Where the heck does that entire -(3a^2)(sin(a^3 + x^3)) term go!?

 

[rocomath]

Hence, dy/dx not dy/dx + dy/da!

 

[illjazz]

Hence, dy/dx not dy/dx + dy/da!

Ah! Ok that makes sense on an intuitive level.. can you expand that a bit though? Does a simply turn into 0 and thus kill that whole term containing a as a factor?

Thanks!

Edit:
By the way.. I stay far, far away from the dy/dx notation wherever possible. I just don't quite understand it.. for instance, I cringe when I see this in my book:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

if

$$\frac{dx}{dt}\neq0$$

for example.. I much prefer something like

F'(x) = f'(g(x)) * g'(x).

Granted, the two examples are not equivalent for the purposes of comparison.. but it gets the point across. I suppose that explains why Leibnitz' notation throws me off :/

 

[rocomath]

$$y'=\frac{dy}{dx}$$

Means to take the derivative of y with respects to x. Basically, only take the derivative of terms that include x.

$$y=ax+a$$

You could easily mistake this as product rule, but it's not! (but you could if you wanted to) Anything that does not have an x variable associated with it, is a constant.

$$\frac{dy}{dx}=a$$

or product rule (pointless)

$$\frac{dy}{dx}=a\frac{dy}{dx}x+x\frac{dy}{dx}a+\frac{dy}{dx}a=a\cdot1+x\cdot0+0=a$$

$$y'=\frac{dy}{dt}$$

Simply means to take the derivative of y with respects to t or time. You can use any variable you want.

 

[HallsofIvy]

Homework Statement

Find the derivative of the function y = cos(a^3 + x^3).

Homework Equations

- Chain rule
- Power rule

The Attempt at a Solution

This is driving me insane.. so here's what I have.

y = cos(a^3 + x^3)

y' = -sin(a^3 + x^3) * d/dx (a^3 + x^3)

y' = -sin(a^3 + x^3) * (3a^2 + 3x^2)

No. d/dx(a^3+ x^3)= 3x^2. a^3 is a constant

y' = -(3a^2)(sin(a^3 + x^3)) - (3x^2)(sin(a^3 + x^3))

I would like to think that that's where this ends.. but my book claims the result is

y' = -(3x^2)(sin(a^3 + x^3))

How? Where the heck does that entire -(3a^2)(sin(a^3 + x^3)) term go!?

 

[illjazz]

Sweet, that definitely helps :)

While we're here.. is it against the rules or anything to ask a second question about a new problem in the same thread? Seems like a bit of a waste to start a new thread just for this next one..

Here goes. If this is bad and frowned upon, let me know! (I understand, obviously, that as a general rule you don't want to do this.. but this one's so short I figure "what the hell"):

$$y=xe^{-x^2}$$

I know the solution already, of course, but I'd appreciate a step by step "walkthrough".

I got as far as this:

$$y=xe^{-x^2}$$

$$y'=x\frac{d}{dx}e^{-x^2}+e^{-x^2}\frac{d}{dx}x$$

$$=xe^{-x^2}+e^{-x^2}$$

Also, can someone tell me why I can't seem to make a new line when writing tex? "\\" is supposed to do it.. but does not :/

Oh btw.. the reason for the last step above was the assumption that the following is true:

$$\frac{d}{dx}e^x=e^x$$

 

[kylera]

$$\frac{d}{dx}e^x=e^x$$ is true because the derivative of x is 1 (by power rule = derivative of x is 1 * x^(1-1)). If, in place of x, there was a function based on x (for example, 2x + 1), you would have e^(2x+1) multiplied by the derivative of the function, resulting in 2e^(2x+1) by the Chain Rule.

Having said that, there is a flaw in your walkthrough. The derivative of e^(-x^2) would be, by the combination of the Chain Rule and the Power Rule, would be -2xe^(-x^2). Hence, your final answer should come out to $$-2x^{2}e^{-x^2}+e^{-x^2}$$.

 

[HallsofIvy]

Sweet, that definitely helps :)

While we're here.. is it against the rules or anything to ask a second question about a new problem in the same thread? Seems like a bit of a waste to start a new thread just for this next one..

It's not "against the rules", but many people will not look at a thread when they know the initial question has been answered- so you may be ruducing the number of people who see your new question.

Here goes. If this is bad and frowned upon, let me know! (I understand, obviously, that as a general rule you don't want to do this.. but this one's so short I figure "what the hell"):

$$y=xe^{-x^2}$$

I know the solution already, of course, but I'd appreciate a step by step "walkthrough".

I got as far as this:

$$y=xe^{-x^2}$$

$$y'=x\frac{d}{dx}e^{-x^2}+e^{-x^2}\frac{d}{dx}x$$

$$=xe^{-x^2}+e^{-x^2}$$

Also, can someone tell me why I can't seem to make a new line when writing tex? "\\" is supposed to do it.. but does not :/

Oh btw.. the reason for the last step above was the assumption that the following is true:

$$\frac{d}{dx}e^x=e^x$$

Yes, that last line is true. But it is NOT true that
$${d e^{x^2}}{dx}= e^{-x^2}$$
You forgot the chain rule: multiply by the derivative of -x².

 

[vertciel]

Hello there:

Here is my solution, which matches your textbook's.

$$y = \cos(a^3 + x^3)$$

Let: $$u = a^3 + x^3$$

The Chain Rule: $$h'(x) = f'(g(x)) \times g'(x)$$

Breaking the original equation down:

$$f(x) = \cos u$$, $$f'(x) = -\sin u$$

$$g(x) = u = a^3 + x^3$$, $$g'(x) = 3x^2$$ (From the Product Rule)

Note: $$3a^2$$ is actually a constant due to the term $$a$$. Recall that the derivative of any constant is 0.)

Compiling each function:

$$h'(x) = f'(g(x)) \times g'(x)$$

$$h'(x) = f'(u) x 3x^2$$

$$h'(x) = -\sin(a^3 + x^3)(3x^2)$$

$$h'(x) = -3x^2 \sin(a^3 + x^3)$$

Hope this helps.

 

[illjazz]

$$\frac{d}{dx}e^x=e^x$$ is true because the derivative of x is 1 (by power rule = derivative of x is 1 * x^(1-1)). If, in place of x, there was a function based on x (for example, 2x + 1), you would have e^(2x+1) multiplied by the derivative of the function, resulting in 2e^(2x+1) by the Chain Rule.

Having said that, there is a flaw in your walkthrough. The derivative of e^(-x^2) would be, by the combination of the Chain Rule and the Power Rule, would be -2xe^(-x^2). Hence, your final answer should come out to $$-2x^{2}e^{-x^2}+e^{-x^2}$$.

Thanks everyone. Your posts helped a lot! What threw me off, I believe, is that e's exponent had itself an exponent.. but what I need to remember specifically is the bold part quoted above!

Thanks again :)

 

[adartsesirhc]

I know this is probably really late, but I'd like to point out to rocomath his/her mistake in using the Leibniz notation: you should have used $$\frac{d}{dx}$$, not $$\frac{dy}{dx}$$. The first one means "take the derivative of" and the second means "the derivative of y with respect to x is".

Just being nitpicky. =]

 

[illjazz]

I know this is probably really late, but I'd like to point out to rocomath his/her mistake in using the Leibniz notation: you should have used $$\frac{d}{dx}$$, not $$\frac{dy}{dx}$$. The first one means "take the derivative of" and the second means "the derivative of y with respect to x is".

Just being nitpicky. =]

And that actually helped me understand Leibniz a bit better! Thanks :)