What is the derivative of 6(5x^2+2sqrt(x^4+5)) ?

The derivative of 6(5x^2+2sqrt(x^4+5)) is (6(20x^5+4x^3+50x))/(sqrt(x^4+5))

What is the first derivative of 6(5x^2+2sqrt(x^4+5)) ?

The first derivative of 6(5x^2+2sqrt(x^4+5)) is (6(20x^5+4x^3+50x))/(sqrt(x^4+5))

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\frac{d}{d x} (6 (5 x^{2} + 2) x^{4} + 5)

\frac{6 ( 20 x ^{5} + 4 x ^{3} + 50 x )}{x ^{4} + 5}

Solution steps

\frac{d}{d x} (6 (5 x^{2} + 2) x^{4} + 5)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d x} ((5 x^{2} + 2) x^{4} + 5)

Apply the Product Rule:

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

f = 5 x^{2} + 2, g = x^{4} + 5

= 6 (\frac{d}{d x} (5 x^{2} + 2) x^{4} + 5 + \frac{d}{d x} (x^{4} + 5) (5 x^{2} + 2))

\frac{d}{d x} (5 x^{2} + 2) = 10 x

\frac{d}{d x} (x^{4} + 5) = \frac{2 x ^{3}}{x ^{4} + 5}

= 6 (10 x x^{4} + 5 + \frac{2 x ^{3}}{x ^{4} + 5} (5 x^{2} + 2))

Simplify

6 (10 x x^{4} + 5 + \frac{2 x ^{3}}{x ^{4} + 5} (5 x^{2} + 2)) : \frac{6 ( 20 x ^{5} + 4 x ^{3} + 50 x )}{x ^{4} + 5}

= \frac{6 ( 20 x ^{5} + 4 x ^{3} + 50 x )}{x ^{4} + 5}

x^{2} + 4, - x^{2} - 1, - 2, 3

\int x^{4} e^{- x^{5}} d x

\int \frac{3 x}{2 x ^{2} + 5} d x

\frac{\partial}{\partial x} (cos (- t + y + e^{x}))

\int x^{5} e^{x^{2}} d x

What is the derivative of 6(5x^2+2sqrt(x^4+5)) ?
The derivative of 6(5x^2+2sqrt(x^4+5)) is (6(20x^5+4x^3+50x))/(sqrt(x^4+5))
What is the first derivative of 6(5x^2+2sqrt(x^4+5)) ?
The first derivative of 6(5x^2+2sqrt(x^4+5)) is (6(20x^5+4x^3+50x))/(sqrt(x^4+5))