What is (\partial)/(\partial x)(3sin^2(x)cos(x)) ?

The answer to (\partial)/(\partial x)(3sin^2(x)cos(x)) is 3(sin(2x)cos(x)-sin^3(x))

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(\partial)/(\partial x)(3sin^2(x)cos(x))

\frac{\partial}{\partial x} (3 sin^{2} (x) cos (x))

3 (sin (2 x) cos (x) - sin^{3} (x))

Solution steps

\frac{\partial}{\partial x} (3 sin^{2} (x) cos (x))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{\partial}{\partial x} (sin^{2} (x) cos (x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = sin^{2} (x), g = cos (x)

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

\frac{\partial}{\partial x} (sin^{2} (x)) = sin (2 x)

\frac{\partial}{\partial x} (cos (x)) = - sin (x)

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

Simplify

3 (sin (2 x) cos (x) + (- sin (x)) sin^{2} (x)) : 3 (sin (2 x) cos (x) - sin^{3} (x))

= 3 (sin (2 x) cos (x) - sin^{3} (x))

\int y sin (x) d x

\int sec (x) (tan (x) + sec (x)) d x

\int \frac{1}{1 + e ^{2 x}} d x

y^{^{'}} = a y + b e^{- b}

\int \frac{e ^{x}}{( e ^{2 x} + 1 ) ( e ^{x} - 1 )} d x

What is (\partial)/(\partial x)(3sin^2(x)cos(x)) ?
The answer to (\partial)/(\partial x)(3sin^2(x)cos(x)) is 3(sin(2x)cos(x)-sin^3(x))