Question

derivative of sqrt x + (1/4) sin (2x)^2

Answer 1

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

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)

Answer 3

hint : derivative of sqrt(f(x)) = 1/(f'(x)*sqrt(f(x)) and
derivative of sin(g(x))=g'(x)*cos(g(x))

Answer 4

could you step by step for the solution (the square is only on the 2x)

Answer 5

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)

Answer 6

@kelly226 is right

Answer 7

is the the (2x)^2 becoming (2x^2) product rule?

Answer 8

i dont think so...

Answer 9

its use chain rule...
derivative of (2x)^2 is 2(2x)*2

Answer 10

wouldnt the derivative of (2x)^2 be 4x then
according to chain rule= 2(2x)^2-1 (2)

Answer 11

2(2x)*2=8x not 4x

Answer 12

but, if we look the first question there is a coefficien of sin (2x)^2 (is 1/4), so multiply it by derivative of sin (2x)^2

Answer 13

therefore, 1/4*2(2x)*2*cos(2x)^2 = 2xcos(2x)^2

Answer 14

I understand the 2xcos part of the solution. But I dont understand why the original (2x)^2 remains

Answer 15

because derivative of sin(f(x) is f'(x)*cos(f(x))

Answer 16

oh okay thank you very much

Answer 17

wellcome :)