What is the derivative of sin^2(2x^3+2x^2) ?

The derivative of sin^2(2x^3+2x^2) is sin(2(2x^2+2x^3))(4x+6x^2)

What is the first derivative of sin^2(2x^3+2x^2) ?

The first derivative of sin^2(2x^3+2x^2) is sin(2(2x^2+2x^3))(4x+6x^2)

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\frac{d}{d x} (sin^{2} (2 x^{3} + 2 x^{2}))

sin (2 (2 x^{2} + 2 x^{3})) (4 x + 6 x^{2})

Solution steps

\frac{d}{d x} (sin^{2} (2 x^{3} + 2 x^{2}))

Apply the chain rule:

2 sin (2 x^{3} + 2 x^{2}) \frac{d}{d x} (sin (2 x^{3} + 2 x^{2}))

= 2 sin (2 x^{3} + 2 x^{2}) \frac{d}{d x} (sin (2 x^{3} + 2 x^{2}))

\frac{d}{d x} (sin (2 x^{3} + 2 x^{2})) = cos (2 x^{3} + 2 x^{2}) (6 x^{2} + 4 x)

= 2 sin (2 x^{3} + 2 x^{2}) cos (2 x^{3} + 2 x^{2}) (6 x^{2} + 4 x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= sin (2 (2 x^{2} + 2 x^{3})) (4 x + 6 x^{2})

\frac{\partial}{\partial x} (6 y^{2} - 4 y^{3} - 3 x^{2} + 6 x y)

f (x) = \frac{6}{5 x + 2}, at x = - 1

f (x) = arctan (x)

\int \frac{1}{sin ( θ )} d θ

\int \frac{1}{3 x - 3} d x

What is the derivative of sin^2(2x^3+2x^2) ?
The derivative of sin^2(2x^3+2x^2) is sin(2(2x^2+2x^3))(4x+6x^2)
What is the first derivative of sin^2(2x^3+2x^2) ?
The first derivative of sin^2(2x^3+2x^2) is sin(2(2x^2+2x^3))(4x+6x^2)