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- anonymous

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)

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- anonymous

- schrodinger

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- anonymous

What do you guys think?

- anonymous

wrong, way off

- anonymous

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

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- anonymous

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

- anonymous

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

- anonymous

=sin2x
du/dx = 2cos(2x)
v= (3x+7)^5
dv/dx = 5 (3x+7)^4 * 3 = 15 (3x+7)^4
now product rule

- anonymous

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

- anonymous

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

- anonymous

thats copied from the guide

- anonymous

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

- anonymous

Thanks for the help

- anonymous

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

- anonymous

Isn't the derivative of cos = -sin?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I may just be really really confused now

- anonymous

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

- anonymous

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 )

- anonymous

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

- anonymous

ok i caught it that time, thanks for the help

- anonymous

you are welcome

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