Question

Using the product rule to differentiate the following:
Y = 8sin(5x)loge(5x)

Answer 1

\[(fg)'=f'g+g'f\] with \[f(x)=\sin(5x), g(x)=\ln(5x)\] so the first thing you need to find is \(f'(x)\) and \(g'(x)\) . both require the chain rule

for the first one since \(\frac{d}{dx}\sin(x)=\cos(x)\) you have \(\frac{d}{dx}\sin(5x)=5\cos(5x)\)

Answer 2

so let y=u*v
and let u=8sin(5x) v=loge(5x)
differentiate your 'u' to get u' and v to get v'

then the formula for the derivative = v*u' + u*v'

i'm a little unclear as to whether it is log(5x) or log(e^5x)

if you could clarify then i can confirm the answer you get

Answer 3

I'm here but I was writing down my own answer

Answer 4

This was my answer : dy/dx= 8sin5x/x + 8cos5xloge5x

Answer 5

Did I get it right?

Answer 6

looks good to me

Answer 7

oops no, hold the phone!

Answer 8

this part 8cos5xloge5x is wrong

Answer 9

Because in the textbook the answer was dy/dx= 8sin5x/x + 40cos5xloge5x

Answer 10

if \(f(x)=\sin(5x)\) then \(f'(x)=5\cos(5x)\) so you are missing a factor of 5 in that part

Answer 11

it is easy to forget the chain rule for something like \(\sin(5x)\) but you need a factor of 5 out front

Answer 12

I almost got it right :) but thanks for the help

Answer 13

you don't see the factor in the derivative of \(\ln(5x)\) because \(\ln(5x)=\ln(5)+\ln(x)\) and \(\ln(5)\) is just a constant

Answer 14

Yes Chad that answer is correct

Answer 15

dy/dx= 8sin5x/x + 40cos5xloge5x is correct :)