find derivative: y=sqrt(x) + (1/4) sin2x^2?

Question

ok im trying to write these without being confusing.
its a lot though hahah.
thanks again!!!

Answer 1

I'm not clear on the last part.

is that
sin (2x)^2
or
(sin 2x)^2

Please update and I'll check back.
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If the square is only on the 2x...

y = sqrt x + (1/4)sin (2x)^2
= sqrt x + (1/4) sin (4x^2)

y' = (1/2)(1 / sqrt x) + (1/4)[cos (4x^2)](4*2x)
= (1 / 2 sqrt x) + (2x)cos (4x^2)
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If the square is meant for the sin...

y = sqrt x + (1/4)[sin 2x]^2

y' = (1/2)(1 / sqrt x) + (1/4)(2)(sin 2x)(cos 2x)(2)
= (1 / 2 sqrt x) + (sin 2x)(cos 2x)

Answer 2

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Answer 3

y=sqrt(x) + (1/4) sin2x^2?
dy=2x+[1/4(-cos2x^2)2(2x)]
=2x-xcos2x^2

Answer 4

y ' = 1 / (2 √x ) + 1/4 cos 2x² (4x)
y ' = 1 / (2 √x ) + x cos 2x²

Answer 5

y=sqrt(x) + (1/4) sin2x^2
dy/dx=d sqrt(x)/dx+1/4d sin2x^2/dx
=1/(2sqrt(x)+1/4*2sin2x dsin2x/dx
=1/(2sqrt(x)+1/2sin2x cos2xd2x/dx
=1/(2sqrt(x)+1/2*2sin2x cos2x
=1/(2sqrt(x)+sin2x cos2x

Answer 6

y = x^(1/2) + (1/4) [ sin 2x ] ²
dy/dx
(1/2) x^(-1/2) + (2/4) [ sin 2x ] 2 cos 2x
1 / [ 2 x^(1/2) ] + sin 2x cos 2x
1 / [ 2 x^(1/2) ] + (1/2) sin 4x

Answer 7

d/dx(sqrt(x)+1/4sin2x^2)
=d/dx(x^1/2)+d/dx(1/4sin2x^2)
=1/2x^(1/2-1)+1/4d/dx(sin2x^2)
=1/2x^-1/2+1/4d/dz(sin z)*dz/dx [ assuming z= 2x^2]
=1/(2x^1/2)+1/4 cos z d/dx(2x^2)
=1/2sqrt(x)+1/4cos2x^2 {2d/dx(x^2)}
=1/2sqrt(x)+1/4*2 cos(2x^2 ) (2x^1)
=1/2sqrt(x)+1/2*2 cos(2x^2 ) (x)
=1/2sqrt(x)+xcos(2x^2 )

Hope this will clear ur doubts

Answer 8

f(x) = sqrt(x) + [sin(2x)^2]/4

// for ([sin(2x)^2]/4)': bottom·deriv of top - top·deriv. of bottom
// all over bottom squared, and [sin(2x)]' = 2cos(2x)

f'(x) = 1/2 (x)^(1/2 - 1) + (4[ 4sin(2x)cos(2x)] - sin(2x)^2·0)/16

f'(x) = 1/[2·sqrt(x)] + 16sin(2x)cos(2x)/16

f'(x) = 1/[2·sqrt(x)] + sin(2x)cos(2x)

// Hope this helps.