Question

if i am looking for the derivative of
f(x)=sin2x(3x+7)^5 would i come out with
f'(x)=5(sin2x)^4(cos2)(6x+21)

Answer 1

What do you guys think?

Answer 2

wrong, way off

Answer 3

there are many chain rules here...make sure you apply them

Answer 4

and i'm assuming that's sin(2x) (3x+7)^5?

Answer 5

review your product rules...and go slowly =)

Answer 6

=sin2x

du/dx = 2cos(2x)

v= (3x+7)^5
dv/dx = 5 (3x+7)^4 * 3 = 15 (3x+7)^4

now product rule

Answer 7

Find the derivative of the function f(x) = sin 2x(3x + 7)^5

Answer 8

f'(x) = 2cos(2x)(3x+7)^5 + 15sin(2x) (3x+7)^4

Answer 9

thats copied from the guide

Answer 10

what elec wrote is correct...so please do it yourself and compare with what he wrote...there's no point copying the answer...make sure you know how he derived it

Answer 11

Thanks for the help

Answer 12

np, let me know if something seems fishy...it's better to do the leg work now....or you'll regret it later =)

Answer 13

Isn't the derivative of cos = -sin?

Answer 14

it is...but you are looking for the derivative of sine which is cosine

Answer 15

so...
d/dx sin(x) = cos(x)
d/dx cos(x) = -sin(x)

Answer 16

ok i see that now thanks, can you break down how he got the 15sin it looks to me that it would be 10sin(2x)(3x+7)^4

Answer 17

I may just be really really confused now

Answer 18

careful there
you shouldn't be doing a chain rule on the sin(2x) ...you need to do it on the (3x+7)...so actually...you get 15sin(2x)(3x+7)^4

Answer 19

it might help to break up your function...
f(x) = sin(2x) * (3x + 7)^5
= u(x) * v(x)
f'(x) = u'(x)*v(x) + v'(x)*u(x)
try to take it from here ( I let u(x) = sin(2x) and v(x) = (3x+7)^5 )

Answer 20

f'(x)=2cos2*(3x+7)^5+15x*sin(2x)(3x+7)^4

Answer 21

ok i caught it that time, thanks for the help

Answer 22

you are welcome