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

The derivative of y=(x^4+2)^2(x^3+4)^4 is 8x^3(x^4+2)(x^3+4)^4+12x^2(x^3+4)^3(x^4+2)^2

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

The first derivative of y=(x^4+2)^2(x^3+4)^4 is 8x^3(x^4+2)(x^3+4)^4+12x^2(x^3+4)^3(x^4+2)^2

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derivative of

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

8 x^{3} (x^{4} + 2) (x^{3} + 4)^{4} + 12 x^{2} (x^{3} + 4)^{3} (x^{4} + 2)^{2}

Solution steps

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

Apply the Product Rule:

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

f = (x^{4} + 2)^{2}, g = (x^{3} + 4)^{4}

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

\frac{d}{d x} ((x^{4} + 2)^{2}) = 8 x^{3} (x^{4} + 2)

\frac{d}{d x} ((x^{3} + 4)^{4}) = 12 x^{2} (x^{3} + 4)^{3}

= 8 x^{3} (x^{4} + 2) (x^{3} + 4)^{4} + 12 x^{2} (x^{3} + 4)^{3} (x^{4} + 2)^{2}

\int - 3 x^{2} sin (π x) d x

y^{^{''}} - 2 y^{^{'}} - 15 y = 3

\frac{\partial}{\partial x} (x^{\frac{1}{2}} \cdot y)

\frac{d}{d x} ((x - \frac{2}{x})^{4})

f (x) = 2 x^{2}, at x = 0

What is the derivative of y=(x^4+2)^2(x^3+4)^4 ?
The derivative of y=(x^4+2)^2(x^3+4)^4 is 8x^3(x^4+2)(x^3+4)^4+12x^2(x^3+4)^3(x^4+2)^2
What is the first derivative of y=(x^4+2)^2(x^3+4)^4 ?
The first derivative of y=(x^4+2)^2(x^3+4)^4 is 8x^3(x^4+2)(x^3+4)^4+12x^2(x^3+4)^3(x^4+2)^2