What is the derivative of 4x^4y^3+cos(x^2) ?

The derivative of 4x^4y^3+cos(x^2) is 4(4x^3y^3+3y^2(dy)/(dx)x^4)-sin(x^2)*2x

What is the first derivative of 4x^4y^3+cos(x^2) ?

The first derivative of 4x^4y^3+cos(x^2) is 4(4x^3y^3+3y^2(dy)/(dx)x^4)-sin(x^2)*2x

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

4 (4 x^{3} y^{3} + 3 y^{2} \frac{d y}{d x} x^{4}) - sin (x^{2}) \cdot 2 x

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (4 x^{4} y^{3}) = 4 (4 x^{3} y^{3} + 3 y^{2} \frac{d y}{d x} x^{4})

\frac{d}{d x} (cos (x^{2})) = - sin (x^{2}) \cdot 2 x

= 4 (4 x^{3} y^{3} + 3 y^{2} \frac{d y}{d x} x^{4}) - sin (x^{2}) \cdot 2 x

\int x^{2 n} d x

\frac{d}{d x} (\frac{x ^{4}}{8})

\frac{\partial}{\partial x} (e^{4 (e^{x} - 1)})

x \to 0 lim ((1 - 6 x)^{\frac{6}{x}})

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

What is the derivative of 4x^4y^3+cos(x^2) ?
The derivative of 4x^4y^3+cos(x^2) is 4(4x^3y^3+3y^2(dy)/(dx)x^4)-sin(x^2)*2x
What is the first derivative of 4x^4y^3+cos(x^2) ?
The first derivative of 4x^4y^3+cos(x^2) is 4(4x^3y^3+3y^2(dy)/(dx)x^4)-sin(x^2)*2x