Help/explanation for some calc stuff

[Whorax]

So I just picked up on the second half of a calc class that I should've taken a year ago. Literally.

Anyways, I had an assignment where I couldn't figure out two questions. I need to find the derivatives of:

 

y = e^9x cos(x)

I ended up figuring out that y' = 9e^(9x cos(x)) (cos(x)-xsin(x)) but I forget how/why the 9 goes down before the e and the product rule stuff doesn't stay in the exponent. Is that because of the chain rule? Why isn't the answer just e^(9cos(x)-xsin(x)) or e^(9x cos(x)) (9cos(x)-xsin(x))? If you can't tell I'm forgetting quite a lot of rules I should know here.

 

and

 

g(t) = 1/(t³ + 5)⁴

This one I would think you'd just do the quotient rule, but unless I'm getting some calculation wrong that I'm just missing that isn't the case.

 

Thanks for any help.

 

Edit: now that I already started this, can someone check my derivative of y = ln(|1-x-6x²|) as well? If memory serves, y' = 1/(6x^2+x+1) * 12x+1, which can then be simplified into 12x+1/(6x^2+x+1), correct? According to the website that I'm using it should be 12x+1/(6x^2+x-1). What am I missing that I'm not getting the minus 1?

[79wjd]

The derivative of e^n is eⁿ * (derivative of n). So: 

derivative of 9x cos(x) = first term * derivative of second term + second term * derivative of the first term --> 

9x*(-sin(x)) + cos(x) * 9 --> -9xsin(x) + 9cos(x)

 

y = e^9xcos(x)

y' = e^9xcos(x) * -9sin(x)+9cos(x) 

 

 

I prefer dealing with the multiplication rule (as I find it easier to keep track of in my head), so:

g(t) = (t³+5)^-4 

g'(t) = -4(t³+5)^-5 * (3t²)

[19_blackie_73]

when making y':

1.) everthing in e^x stays the same

2.) that's whats in e^x (here: 9x*cosx) will be differentiated and multiplied with the e^x term

3.) because 9x*cosx is a product (u*v), it will be differentiated u'*v (9*cosx) + u*v' (9x* (-sinx)) = 9 cosx-9x sinx

4.) because 9 appears in both terms, you can pull it in front of the whole term: 9 (cosx - x sinx)

5.) the term from 4.) will be multiplied with the e^x thingy => y'(x)=9 (cosx - x sinx) e^(9x cosx)

 

g(t):

you can write it as: g(t)=1*(t³+5)^(-4)

you could do the quotient rule, but it's slower than making the normal differentiation: -4*(t³+5 )^(-5) * 3t²

the t² appears because it is a linked function: y^(-4) with y = t³+5

 

EDIT: sry for maybe not the correct english expressions, it's hard to translate it form german to englisch:D

differentiation means dy/dx=y'

@Whorax

[79wjd]

For your Edit: 

 

1/(1-x-6x²) * (12x-1) --> (12x-1)/(1-x-6x²) --> (-12x-1) / ((3x-1)(-2x-1)) --> (-12x-1) / ((3x-1)*-1(2x+1)) --> (12x+1) / ((3x-1)(2x+1) --> (12x+1)/(6x²+x-1)

[Whorax]

-snip-

Thank you. I suppose bringing everything up from the denominator does make it easier. I'm still going to try to work through it with the quotient rule just for the sake of doing it that way, but I now have a correct answer so thanks.

 

-snip

Thanks. I think I gotcha on why the 9 can go in front now. Makes a bit more sense!

[79wjd]

Thank you. I suppose bringing everything up from the denominator does make it easier. I'm still going to try to work through it with the quotient rule just for the sake of doing it that way, but I now have a correct answer so thanks.

 

Thanks. I think I gotcha on why the 9 can go in front now. Makes a bit more sense!

g'(t) = (bottom term * derivative of top term - top term * derivative of bottom term) / bottom term squared

g(t) = 1/(t3 + 5)4

g'(t) = - 4(t³+5)³(3t²) / (t³+5)⁸