Recent content by Vanessa13

Thank you. Would you mind if I ever messaged you with a question now and then? As you can probably tell, Calculus isn't exactly my strong suit.

v' = -2sin(2t)sin(6t)*7 + cos(2t)*6cos(6t)*7 + 0 v' = -14sin(2t)sin(6t) + 42cos(2t)cos(6t) [Took 14 out...] v' = 14(-sin(2t)sin(6t)) + 3(cos(2t)cos(6t)) [Do I continue to add them?]

I feel really ignorant now. derivative of cos(2t) = -2sin(2t) derivative of sin(6t) = 6cos(6t) derivative of 7 = 0

I see that now. Here goes, third times a charm (hopefully)... v' = -7sin2sin6t + 7cos2tcos6 + 0 * cos2tsin6t = -7sin2sin6t + 7cos2tcos6 (then you take out a 7, right?) v' = 7(-sin2sin6t)(cos2tcos6) I'm not sure if I'm still supposed to be adding them [(-sin2sin6t) +...

Okay, I did it this way and it makes a lot more sense. f'(x)= -sin2tsin6t + 7cos2tcos6t I think that's all I can do with it, right?

That's probably true... Thank you for being my savior! :) It doesn't have parenthesis around them in my book, so I'm not sure at all... That's why I was so confused.

I got confused again :( f(x) = 7cos2tsin6t I separated them like this: u= 7cos v =2t y =sin6t u'= -sin v'= 2 y'= cos6 So far my derivative looks like this: f'(x) = -sin2tsin6t + 7cos2sin6t + 7cos2tcos6 This is the farthest it can be...

:) Thank you so much!

I always make basic errors. Alright. So with those fixed, this is what my simplified derivative came out to be: 6x2(x3+4)5(x3+6)3(5x3+26) Better? Also, THANK YOU!

I'm beginning to understand. If you don't mind, I'm going to show you what I did step by step... Hopefully I do it right. 6x2(x3+4)5(x3+6)3* 3(x3+6) + 12(x3+4) = 6x2(x3+4)5(x3+6)3* (3x3+12+12x3+48) = 6x2(x3+4)5(x3+6)3*(15x3+60) Did I do the simplifying right?

Homework Statement f(x) = (x^3+6)^4 * (x^3+4)^6 Homework Equations f'(x) = (x^3+6)^4*18x^2(x^3+4)^5 + (x^3+4)^6*12x^2(x^3+6)^3 The Attempt at a Solution Obviously this expression can be simplified, but I haven't the slightest clue on how to do this. I would really appreciate...