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Popular >

derivative of (3cos(2x+sin(3x)+4)^2)

Solution

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

Solution

2 (3 cos (2 x) + sin (3 x) + 4) (3 cos (3 x) - 6 sin (2 x))

Solution steps

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

Apply the chain rule:

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

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

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

= 2 (3 cos (2 x) + sin (3 x) + 4) (3 cos (3 x) - 6 sin (2 x))

Popular Examples

y^'-4y=10sin(2t),y(0)=0

y^{^{'}} - 4 y = 10 sin (2 t), y (0) = 0

derivative of f(x)=-3/(e^{3x)}derivative of

f (x) = - \frac{3}{e ^{3 x}}

limit as x approaches-3 of x^2+4x+1

x \to - 3 lim (x^{2} + 4 x + 1)

integral of 1/(x(4-x))

\int \frac{1}{x ( 4 - x )} d x

(\partial)/(\partial y)(ye^{2xy}+x)

\frac{\partial}{\partial y} (y e^{2 x y} + x)

Frequently Asked Questions (FAQ)

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