Question

find the first derivative with respect to x of y= 3x/8 + (sinxcosx)/8 +(cos^3 xsinx)/4
a.sin^4 x
b.-sin^4 x
c. cos^4 x
d. -cos^4 x

Answer 1

y= 3x/8 + (sinxcosx)/8 +(cos^3 xsinx)/4

y' = 3/8 + 1/8 * (cos^2(x) - sin^2(x)) + 1/4 * (-3cos^2(x)*sin(x) + cos^4(x) ) =

= 3/8 + 1/8 * (cos^2(x) - sin^2(x)) + 1/4 * (-3cos^2(x)*sin(x) + cos^4(x) ) =

= 3/8 + 1/8 * (1-2*sin^2(x)) + 1/4 * (-3(1-sin^2(x))*sin(x) + cos^4(x) ) =

= 3/8 + 1/8 * (1-2*sin^2(x)) + 1/4 * (-3(1-sin^2(x))*sin(x) + (1-sin^2(x))^2 ) =
= 3/8 + 1/8 * (1-2*sin^2(x)) + 1/4 * (-3(1-sin^2(x))*sin(x) + (1-sin^2(x))^2 ) =

= 3/8 + 1/8 * (1-2*sin^2(x)) + 1/4 * (-3(1-sin^2(x))*sin(x) + 1-2sin^2(x)+sin^4(x) ) =

= 3/8 + 1/8 + 1/4 - 3/4*sin(x) + (1/4-2/4)*sin^2(x) + 3/4*sin^3(x) + 1/4*sin^4(x) =
.= 6/8 - 3/4*sin(x) -1/4*sin^2(x) + 3/4*sin^3(x) + 1/4*sin^4(x) =

= 6/8 - 3/4*sin(x)(1-sin^2(x)) -1/4*sin^2(x) (1-sin^2(x)) =

= 6/8 - 1/4*(1-sin^2(x))sin(x) (3 + sin(x))

Answer 2

then ..

Answer 3

i vote for none of the above

Answer 4

http://www.wolframalpha.com/input/?i=3x%2F8+%2B+%28sinxcosx%29%2F8+%2B%28cos^3+xsinx%29%2F4

Answer 5

ok this is the longest part of my h.w i will put E.