What is the derivative of x^2cos(yy^') ?

The derivative of x^2cos(yy^') is 2xcos(y)y^'+(-sin(y)(dy)/(dx)y^'+'cos(y)(dy)/(dx)y^{'-1})x^2

What is the first derivative of x^2cos(yy^') ?

The first derivative of x^2cos(yy^') is 2xcos(y)y^'+(-sin(y)(dy)/(dx)y^'+'cos(y)(dy)/(dx)y^{'-1})x^2

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\frac{d}{d x} (x^{2} cos (y) y^{^{'}})

2 x cos (y) y^{'} + (- sin (y) \frac{d y}{d x} y^{'} +' cos (y) \frac{d y}{d x} y^{' - 1}) x^{2}

Solution steps

\frac{d}{d x} (x^{2} cos (y) y^{^{'}})

Apply the Product Rule:

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

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

= \frac{d}{d x} (x^{2}) cos (y) y^{^{^{'}}} + \frac{d}{d x} (cos (y) y^{^{^{'}}}) x^{2}

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (cos (y) y^{^{'}}) = - sin (y) \frac{d y}{d x} y^{^{'}} +^{'} cos (y) \frac{d y}{d x} y^{^{'} - 1}

= 2 x cos (y) y^{^{^{'}}} + (- sin (y) \frac{d y}{d x} y^{^{^{'}}} +^{^{'}} cos (y) \frac{d y}{d x} y^{^{^{'}} - 1}) x^{2}

What is the derivative of x^2cos(yy^') ?
The derivative of x^2cos(yy^') is 2xcos(y)y^'+(-sin(y)(dy)/(dx)y^'+'cos(y)(dy)/(dx)y^{'-1})x^2
What is the first derivative of x^2cos(yy^') ?
The first derivative of x^2cos(yy^') is 2xcos(y)y^'+(-sin(y)(dy)/(dx)y^'+'cos(y)(dy)/(dx)y^{'-1})x^2