## stamp Group Title [CALCULUS III—CHAIN RULE] Find dw/dt by the chain rule where w = xy cos(z) and x = t, y = t^2, and z = arccos(t). one year ago one year ago

1. stamp Group Title

$w=xy\ cos(z)$$w=t(t^2)\ cos(arccos(t))$$w=t^3t$$w=t^4$$dw/dt=4t^3$ Is that all there is to it or am I mistaken?

2. amistre64 Group Title

by the chain rule eh $w=xy\ cos(z)$ $w(t)=x(t)y(t)\ cos(z(t))$ $w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))+x(t)y(t)\ cos'(z(t))$ $w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))-x(t)y(t)z'(t)\ sin(z(t))$

3. stamp Group Title

So did I goof?

4. stamp Group Title

Or would all that simplify to 4t^3

5. amistre64 Group Title

we can ignore the (t) parts to clean it up $w'=x'y\ cos(z)+xy'\ cos(z)-xyz'\ sin(z)$ $w'=cos(z)(x'y+xy')-xyz'\ sin(z)$ $w'=cos(z)(x'y+xy')-xyz'\ sin(z)$ x = t, y = t^2, and z = arccos(t) x = 1, y = 2t, and z = $$\frac{1}{\sqrt{1+...}}$$ $w'=cos(cos^{-1}z)(y+2t^2)-t^3z'\ sin(cos^{-1}z)$

6. amistre64 Group Title

$w'=z(t^2+2t^2)-t^3z'\ sin(cos^{-1}z)$ $w'=3t^2z-t^3z'\ sin(cos^{-1}z)$ is that looking right so far?

7. stamp Group Title

You forgot your x' and y' and z' on your derivatives but it is ok because I recognize what you did.

8. amistre64 Group Title

it does simplify out to 4t^3

9. stamp Group Title

$w′=cos(z)(x′y+xy′)−xyz′ sin(z)$ x'y = t^2 xy' = 2t x'y + xy' = 2t + t^2 = t(2+t) cos(z) = cos(arccos t) = t t( t(2+t) ) - xyz' sin(z)

10. amistre64 Group Title

provided t not equal -1 or 1 due to that z'

11. stamp Group Title

I am not quite seeing the simplification to 4t^3

12. stamp Group Title

I understand the first part where you showed the formal chain rule of w(t) = x(t) y(t) cos(z(t)) The simplification of it to 4t^3 is still eluding me.

13. stamp Group Title

ps thank you for your assistance so far

14. amistre64 Group Title

i got lost, let me go back to the w' chains :) and sort it back out

15. stamp Group Title

ok

16. amistre64 Group Title

$w'=x'y\cos(z)+xy'\cos(z)+xy~\cos'(z(t))$ $w'=x'y\cos(z)+xy'\cos(z)-xyz'~\sin(z)$ $w'=x'y\cos(z)+xy'\cos(z)-xyz'~\sin(z)$ $x = t,~ y = t^2, ~z = cos^{-1}(t)$$x' = 1, ~y' = 2t, ~z' =(1-t^2)^{(-1/2)}$ $w'=t^2t+t~2t~t-tt^2(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))$ $w'=t^3+2t^3-t^3(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))$ $w'=t^3(3-(1-t^2)^{(-1/2)}~\sin(\cos^{-1}(t))$ the key here is in definine sin(cos^-1 (t)) better

17. amistre64 Group Title

|dw:1361983895329:dw|

18. amistre64 Group Title

we still have to include the caveat of t not equal to -1 or 1

19. stamp Group Title

sin(arccos(t)) = sqrt( 1 - t^2 )

20. amistre64 Group Title

so that reduces to w' = (3-1) t^3 which means i prolly missed a sign in there someplace along the way :)

21. amistre64 Group Title

z' = -(1-t^2)^1/2

22. amistre64 Group Title

i forgot the - part :)

23. stamp Group Title

w' = (3-1)t^3 w' = 2t^3 ?? :(

24. amistre64 Group Title

check out z', its spose to be negative ... not positive

25. stamp Group Title

Ok. I will ponder all of this and do some work on paper to verify everything. Thank you again for your time.

26. amistre64 Group Title

youre welcome

27. stamp Group Title

@amistre64 Here is the derivative of x(t) y(t) cos z(t) http://www.wolframalpha.com/input/?i=derivative+of+x%28t%29y%28t%29+cos%28z%28t%29%29+

28. amistre64 Group Title

yes, but z' is negative so that changes the sign .... and carries it to the end as (3+1) not (3-1)