Question

find the first derivative of y = sin^2(cos2x)

Answer 1

\[y=Sin^2(Cos(2x))=(Sin(Cos(2x))^2\]

Answer 2

my answer is -4sin(cos2x)(sin(2x))

Answer 3

Use the power/chain rule:
\[y=2(Sin(Cos(2x))*(Sin(Cos(2x))'\]\[(Sin(Cos(2x))'=Cos(Cos(2x))*(Cos(2x))'\]\[(Cos(2x))' = -Sin(2x)*(2x)'=-2Sin(2x)\]

Answer 4

Now just put those back together:\[(Sin(Cos(2x))'=-2Cos(Cos(2x))Sin(2x)\]
By the way, the 2nd line in the post above should have been:
\[y'=2(Sin(Cos(2x))*(Sin(Cos(2x))'\]

Answer 5

y=sin^(2)(cos2x)

Find the derivative of the expression.
(d)/(dx) sin^(2)(cos2x)

The chain rule states that the derivative of a composite function (f o g o h) is equal to (f' o g o h)*g’(h)*h’. To find the derivative of sin^(2)(cos2x), find the derivatives of each portion of the function and use the chain rule formula.
(f o g o h)'(x)=(f' o g o h)*g'(h)*h' where

To find the derivative of u^(2), multiply the base (u) by the exponent (2), then subtract 1 from the exponent.
(d)/(du) u^(2)=2u

The derivative of sin(v) is (cos(v)).
(d)/(dv) sin(v)=(cos(v))

Remove the parentheses around the expression cos(v).
(d)/(dv) sin(v)=cos(v)

The chain rule states that the derivative of a composite function (j o k)' is equal to (j' o k)*k'. To find the derivative of cos2x, find the derivatives of each portion of the function and use the chain rule formula.
(d)/(du) cos(u)*(d)/(dx) 2x

The derivative of cos(u) is -sin(u).
(d)/(du) cos(u)=-sin(u)

To find the derivative of 2x, multiply the base (x) by the exponent (1), then subtract 1 from the exponent (1-1=0). Since the exponent is now 0, x is eliminated from the term.
(d)/(dx) 2x=2

Replace the variable u with 2x in the expression.
(d)/(du) cos(u)=-sin((2x))

Replace the variable u with 2x in the expression.
2

Form the derivative by substituting the values for each portion into the chain rule formula.
=-sin((2x))*2

Multiply -sin((2x)) by 2 to get -2sin((2x)).
(d)/(dx) cos2x=-2sin((2x))

Replace the variable v with cos2x in the expression.
(d)/(dv) sin(v)=cos((cos2x))

Form the derivative by substituting the values for each portion into the chain rule formula.
(f o g o h)'(x)=(f' o g o h)*g'(h)*h'=2(sin((cos2x)))*cos((cos2x))*-2sin((2x))

Multiply -4cos((cos2x))sin((2x)) by each term inside the parentheses.
(d)/(dx) sin^(2)(cos2x)=-4sin((cos2x))cos((cos2x))sin((2x))

The derivative of y with respect to x is -4sin((cos2x))cos((cos2x))sin((2x))

Answer 6

\[y'=2(Sin(Cos(2x))∗(Sin(Cos(2x))'=2(Sin(Cos(2x))* −2Cos(Cos(2x))Sin(2x)\]

Answer 7

Thank you so much for your efforts!

Answer 8

\[y' = -4Sin(Cos(2x))Cos(Cos(2x))Sin(2x)\]