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What do you guys think?
wrong, way off
there are many chain rules here...make sure you apply them
and i'm assuming that's sin(2x) (3x+7)^5?
review your product rules...and go slowly =)
=sin2x du/dx = 2cos(2x) v= (3x+7)^5 dv/dx = 5 (3x+7)^4 * 3 = 15 (3x+7)^4 now product rule
Find the derivative of the function f(x) = sin 2x(3x + 7)^5
f'(x) = 2cos(2x)(3x+7)^5 + 15sin(2x) (3x+7)^4
thats copied from the guide
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
Thanks for the help
np, let me know if something seems fishy...it's better to do the leg work now....or you'll regret it later =)
Isn't the derivative of cos = -sin?
it is...but you are looking for the derivative of sine which is cosine
so... d/dx sin(x) = cos(x) d/dx cos(x) = -sin(x)
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
I may just be really really confused now
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
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 )
ok i caught it that time, thanks for the help
you are welcome