What is the derivative of 4x^2sqrt(5x+1) ?

The derivative of 4x^2sqrt(5x+1) is (2(25x^2+4x))/(sqrt(5x+1))

What is the first derivative of 4x^2sqrt(5x+1) ?

The first derivative of 4x^2sqrt(5x+1) is (2(25x^2+4x))/(sqrt(5x+1))

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\frac{d}{d x} (4 x^{2} 5 x + 1)

\frac{2 ( 25 x ^{2} + 4 x )}{5 x + 1}

Solution steps

\frac{d}{d x} (4 x^{2} 5 x + 1)

Take the constant out:

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

= 4 \frac{d}{d x} (x^{2} 5 x + 1)

Apply the Product Rule:

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

f = x^{2}, g = 5 x + 1

= 4 (\frac{d}{d x} (x^{2}) 5 x + 1 + \frac{d}{d x} (5 x + 1) x^{2})

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

\frac{d}{d x} (5 x + 1) = \frac{5}{2 5 x + 1}

= 4 (2 x 5 x + 1 + \frac{5}{2 5 x + 1} x^{2})

Simplify

4 (2 x 5 x + 1 + \frac{5}{2 5 x + 1} x^{2}) : \frac{2 ( 25 x ^{2} + 4 x )}{5 x + 1}

= \frac{2 ( 25 x ^{2} + 4 x )}{5 x + 1}

\int \frac{5}{4 sin ( x ) + 3 cos ( x )} d x

x^{2} y^{^{''}} + 2 x y^{^{'}} - 6 y = 0, y (1) = 0, y^{^{'}} (1) = - 4

\frac{d}{d x} (sin (ln (4) x^{2}))

y = x 400 - x^{2}

What is the derivative of 4x^2sqrt(5x+1) ?
The derivative of 4x^2sqrt(5x+1) is (2(25x^2+4x))/(sqrt(5x+1))
What is the first derivative of 4x^2sqrt(5x+1) ?
The first derivative of 4x^2sqrt(5x+1) is (2(25x^2+4x))/(sqrt(5x+1))