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

The derivative of y=(x^2+2)^2(x^4+4)^4 is 4x(x^2+2)(x^4+4)^4+16x^3(x^4+4)^3(x^2+2)^2

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

The first derivative of y=(x^2+2)^2(x^4+4)^4 is 4x(x^2+2)(x^4+4)^4+16x^3(x^4+4)^3(x^2+2)^2

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derivative of

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

4 x (x^{2} + 2) (x^{4} + 4)^{4} + 16 x^{3} (x^{4} + 4)^{3} (x^{2} + 2)^{2}

Solution steps

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

Apply the Product Rule:

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

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

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

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

\frac{d}{d x} ((x^{4} + 4)^{4}) = 16 x^{3} (x^{4} + 4)^{3}

= 4 x (x^{2} + 2) (x^{4} + 4)^{4} + 16 x^{3} (x^{4} + 4)^{3} (x^{2} + 2)^{2}

\int \frac{1}{y ^{3} - y} d y

h (x) = \frac{7}{x}, at x = 5

\frac{d}{d x} (- \frac{1}{2} e^{- \frac{x}{4}})

x \to 0 + lim (x lo g_{e} (x))

- 8 (64 e^{- 4 x^{2}} x^{3} - 24 e^{- 4 x^{2}} x)

What is the derivative of y=(x^2+2)^2(x^4+4)^4 ?
The derivative of y=(x^2+2)^2(x^4+4)^4 is 4x(x^2+2)(x^4+4)^4+16x^3(x^4+4)^3(x^2+2)^2
What is the first derivative of y=(x^2+2)^2(x^4+4)^4 ?
The first derivative of y=(x^2+2)^2(x^4+4)^4 is 4x(x^2+2)(x^4+4)^4+16x^3(x^4+4)^3(x^2+2)^2