At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

Is a a constant?

are you differentiating with respect to x..?

or a? :P .

Yes @campbell_st. I'm not sure @Dido525

That means a is a constant I assume O_o .

|dw:1352349392745:dw|

let u = a^3 + x^3 du/dx = 3x^2
let y = cos(u) du/dy = -sin (u)
so dy/dx = dy/du * du/dx

I can't see the rest.

One moment

\(\ \Huge -sinx(3a^2+3x^2) \)

then replace u with a^3 + x^3 for the final solution.

nope its
dy/dx = -sin(a^3 + x^3) * 3x^2

@campbell_st How did you get that?

Because a is a constant. The derivative of a constant is 0.

We multiply by the derivative of the inside part when using the chain rule.

and I used the chain rule..

the a^3 x^3 doesn't change

@campbell_st I thought a was a constant then? How does that stay the same?

Here:
|dw:1352349668500:dw|

We DO NOT change the inside.

|dw:1352349758941:dw|

Differentiate that.

With respect to x. Meaning a is a constant. We know the derivative of a constant is 0.

Usually letters a, b, and c are considered constants.

Well there are partial derivatives but you probably won't learn that now.

its just like
(x^2 + 3)^4 dy/dx = 4(x^2 + 3) ^3 *2x
the brackets don't change

you do... thats how you get 3x^2

But the term is still inside the parenthesis? That's really confusing me right now

parentheses*

|dw:1352350103544:dw|

Chain rule says d/dx(cos u)= -sin u (u')

yep and du/dx = u'
you are differentiating u with respect to x

So u= a^3 + x^3

U'= 0 + 3x^2

Ok well u know d/dx of cos x= -sin x right?

The expression inside of the expression on the outside ALWAYS remains. It's just how it is.

\(\ \text{Yes!} \)

"on the outside"?

Okay\(\ \text{...} \)

Angle measure?