Question

find dy/dx
http://prntscr.com/3q49os

Answer 1

I thought @ganeshie8 showed you how to do this already...? I saw him spent quite some time on it, or is there a specific question?

Answer 2

can you start helping from Rule 3 @ganeshie8

Answer 3

I'm not sure of what you mean, but hopefully ganeshie will get back to you :)

Answer 4

we're done wid problem C right ?

Answer 5

For problem D, simplify the given expression first :

\(\large y = (50x^4)(20x^8)\)

Answer 6

Can you ? :)

Answer 7

i m more confused with c can you explain it ones more

Answer 8

heard of `chain rule` before ?

Answer 9

\(\large y = \cos x * \sin x = \dfrac{1}{2} 2\cos x *\sin x = \dfrac{1}{2}\sin (2x) \)


fine with this simplification ?

Answer 10

no can you explain please

Answer 11

\(\large y = \cos x * \sin x = \dfrac{1}{2} 2\cos x *\sin x \)

Answer 12

fine up to here ?

Answer 13

ok cosx* sinx trig property 1/2 (2cosx*sinx)

Answer 14

yes, we use the trig identity next :
\(\large 2\cos x~\sin x = \sin (2x)\)

Answer 15

you need to knw few trig identities before diving into calculus :
http://mathworld.wolfram.com/Double-AngleFormulas.html

Answer 16

So the given expression simplifies to :

\(\large y = \dfrac{1}{2}\sin (2x)\)

differentiate now

Answer 17

\(\large \dfrac{d}{dx}\left(y\right) = \dfrac{d}{dx}\left(\dfrac{1}{2}\sin(2x)\right)\)

Answer 18

ok did we get this
\[d/dx(y)=1/2 \cos (2x) d/dx(2x)\]

Answer 19

Exactly ! thats the chain rule.

whats the derivative of 2x ?

Answer 20

\(\large \dfrac{d}{dx}\left(y\right) = \dfrac{d}{dx}\left(\dfrac{1}{2}\sin(2x)\right)\)


\(\large~~~~~~~~ = \dfrac{1}{2}\dfrac{d}{dx}\left(\sin(2x)\right)\)


\(\large~~~~~~~~ = \dfrac{1}{2}\cos (2x) \dfrac{d}{dx}\left(2x\right)\)

Answer 21

whats the derivative of 2x ?

Answer 22

hey no
thats integration

Answer 23

derivative of 2x is 2

Answer 24

http://www.wolframalpha.com/input/?i=d%2Fdx%28cosx*sinx%29