Help Me Differentiate!

[DeadlyGambit]

I'm having a little trouble with a problem on my calculus homework. I know it isn't that difficult, but I'm not quite sure how to go about it.

Find the derivative of y = sin(x^4-x^3+2x)^5

Simplification isn't necessary, just the calculus.

Heres an example of another problem that I solved from the same homework, just so you know what I'm looking for:

y = [sin(x^2-1)]^3

dy/dx = 3[sin(x^2-1)]^2[cos(x^2-1)](2x)

I think thats correct. If not let me know.

Thank you,
-Matt

[jay_shark]

this problem involves the chain rule twice .

y'= 5*sin(x^4-x^3+2x)^4[cos(x^4-x^3+2x)(4x^3-3x^2+2)

Just let sin(x^4-x^3+2x) = g(x)

Therefore y=g(x)^5 and y' = 5g(x)^4*g'(x)

[jay_shark]

I solved a slightly different problem where the exponent 5 is over sin(x^4-x^3+2x)

But you probably want it solved the way you've written it .

In that case y'=cos(x^4-x^3+2x)^5*5(x^4-x^3+2x)^4(4x^3-3x^2+2)

[DeadlyGambit]

Thanks Jay. How bout my example, was it correct?

[DeadlyGambit]

Crap. Now I'm confused again. So the original answer you gave isn't correct? Or is the original one the same as the 2nd one you gave, just written in a different way? Let me clarify the problem: its the sine of the quantity (x^4-x^3+2x)^5. I don't know if that helps.

[jay_shark]

The first answer I gave is for y= [sin(x^4-x^3+2x)]^5

The second answer I gave is for y= sin (x^4-x^3+2x)^5

You should note the difference between sin(x^2) and (sinx)^2.

[Senor Cardgage]

The original answer he gave assumed that it was (sin(...))^5. The second post would be the solution with sin((...)^5), which I guess is what you want.

The other problem you asked about is correct.

EDIT: Well, I guess he cleared that up. /images/graemlins/smile.gif Never mind. Off to look at graph theory homework...

[armPitt]

graph theory, eh? must have been the easiest A i ever got, but maybe my teacher was easy.

and i guess it may have been an A-, i don't remember. but you shouldn't have trouble in there.