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

The derivative of f(x)=(x^4)sqrt(x^2+2) is (5x^5+8x^3)/(sqrt(x^2+2))

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

The first derivative of f(x)=(x^4)sqrt(x^2+2) is (5x^5+8x^3)/(sqrt(x^2+2))

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derivative of

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

\frac{5 x ^{5} + 8 x ^{3}}{x ^{2} + 2}

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

= 4 x^{3} x^{2} + 2 + \frac{x}{x ^{2} + 2} x^{4}

Simplify

4 x^{3} x^{2} + 2 + \frac{x}{x ^{2} + 2} x^{4} : \frac{5 x ^{5} + 8 x ^{3}}{x ^{2} + 2}

= \frac{5 x ^{5} + 8 x ^{3}}{x ^{2} + 2}

\int \frac{48 x}{( x + 1 ) ^{2} ( 3 x + 1 )} d x

t^{2} y^{^{''}} - t (t + 2) y^{^{'}} + (t + 2) y = 0

y = 4 x^{2} + 3 x (- 3.27)

x \to 2 lim (\frac{( x + 2 )}{x - 2})

\frac{d}{d y} (\frac{1}{3} (y^{2} + 2)^{\frac{3}{2}})

What is the derivative of f(x)=(x^4)sqrt(x^2+2) ?
The derivative of f(x)=(x^4)sqrt(x^2+2) is (5x^5+8x^3)/(sqrt(x^2+2))
What is the first derivative of f(x)=(x^4)sqrt(x^2+2) ?
The first derivative of f(x)=(x^4)sqrt(x^2+2) is (5x^5+8x^3)/(sqrt(x^2+2))