## bii17 3 years ago find dy/dx.. y= 3/8 (x) + 3/8 (sinxcosx) + 1/4 (cos^3 x sinx)

1. bii17

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

2. ali110

3. bii17

the final answer should be cos^4 x

4. hartnn

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

5. bii17

yep.. can u help me @hartnn

6. hartnn

sure, u know the product rule in differentiation ?

7. bii17

yes..

8. hartnn

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

9. bii17

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

10. hartnn

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

11. bii17

$\sin 2x= 2sinxcosx$

12. bii17

am i right?

13. hartnn

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 ?

14. bii17

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

15. hartnn

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

16. bii17

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

17. hartnn

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 ??

18. bii17

there are 3 formulas for cos 2x

19. hartnn

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

20. bii17

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

21. hartnn

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

22. bii17

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

23. bii17

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

24. hartnn

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

25. bii17

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

26. hartnn

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

27. bii17

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

28. hartnn

welcome :)

29. bii17

^^