find dy/dx..
y= 3/8 (x) + 3/8 (sinxcosx) + 1/4 (cos^3 x sinx)

Question

find dy/dx..
y= 3/8 (x) + 3/8 (sinxcosx) + 1/4 (cos^3 x sinx)

anonymous · Answer

$$\frac{ 3 }{ 8 }x + \frac{ 3 }{ 8 }sinxcosx + \frac{ 1 }{ 4 }\cos^3x sinx$$

anonymous · Answer

attachment

anonymous · Answer

the final answer should be cos^4 x

hartnn · Answer

it is indeed cos^4 x
still interested in finding how ?

anonymous · Answer

yep.. can u help me @hartnn

hartnn · Answer

sure, u know the product rule in differentiation ?

anonymous · Answer

yes..

hartnn · Answer

then tell me what u got the derivative of $cos^3x sin x=?$

anonymous · Answer

$$(\cos^3x) (\cos x) + (\sin x)(3)(\cos^2 x)(-\sin x)$$

hartnn · Answer

that is correct :) 
now u know the formula of sin 2x ?

anonymous · Answer

$$\sin 2x= 2sinxcosx$$

anonymous · Answer

am i right?

hartnn · Answer

yes! u see how we can use that to differentiate middle term easily...
3/8 sin x cos x = 3/16 sin2x
after diff. 
6/16 cos2x
= 3/8 cos 2x
ok with this ?

anonymous · Answer

where did u get 6/16 cos 2x??

hartnn · Answer

$\huge \frac{d}{dx}sin2x=2cos2x$

anonymous · Answer

ok.. and then.. what's next??

hartnn · Answer

the first term diff. is just 3/8.
now the 'differentiation' part is over.Lets put it all together and see what we can simplify:
$\large (3/8)+3/8cos2x-(3/4)cos^2xsin^2x+(1/4)cos^4x$
now do u know the formula for cos2x ??

anonymous · Answer

there are 3 formulas for cos 2x

hartnn · Answer

right, lets get everything in terms of cos....so which formula has cos in it ?

anonymous · Answer

cos 2x= cos^2 x - sin^2 x
cos 2x= 2cos^2 x - 1
 these 2 formulas have cos

hartnn · Answer

lets use 2nd one,
where we directly get $1+cos2x=2cos^2x$
 and let use the fact that $sin^2x=1−cos^2x$
so now we have :
$(3/4)cos^2x−(3/4)cos^2x(1−cos^2x)+(1/4)cos^4x$
can u simplify this and tell me what gets cancelled

anonymous · Answer

hmmm...can i cancell both (3/4)cos^2x ???

anonymous · Answer

hmm ok i'll get it ... cos^4 x will be left.

hartnn · Answer

but did u get how i got that step ?
ask if there is any doubt in any step/term.

anonymous · Answer

yes.. I get it..  but about this..  sin^2 x = 1- cos^2 x where din u get it?/

hartnn · Answer

basic trignometric identity
$\huge \color{red}{sin^2x+cos^2x=1}$

anonymous · Answer

ow... Thank You For your help :))

hartnn · Answer

welcome :)

anonymous · Answer

^^