What is the derivative of (2x^3-5(x^2-5ln(x))) ?

The derivative of (2x^3-5(x^2-5ln(x))) is 10x^4-30x^2ln(x)-10x^2-10x+(25)/x

What is the first derivative of (2x^3-5(x^2-5ln(x))) ?

The first derivative of (2x^3-5(x^2-5ln(x))) is 10x^4-30x^2ln(x)-10x^2-10x+(25)/x

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\frac{d}{d x} ((2 x^{3} - 5) (x^{2} - 5 ln (x)))

10 x^{4} - 30 x^{2} ln (x) - 10 x^{2} - 10 x + \frac{25}{x}

Solution steps

\frac{d}{d x} ((2 x^{3} - 5) (x^{2} - 5 ln (x)))

Apply the Product Rule:

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

f = 2 x^{3} - 5, g = x^{2} - 5 ln (x)

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

\frac{d}{d x} (2 x^{3} - 5) = 6 x^{2}

\frac{d}{d x} (x^{2} - 5 ln (x)) = 2 x - \frac{5}{x}

= 6 x^{2} (x^{2} - 5 ln (x)) + (2 x - \frac{5}{x}) (2 x^{3} - 5)

Simplify

6 x^{2} (x^{2} - 5 ln (x)) + (2 x - \frac{5}{x}) (2 x^{3} - 5) : 10 x^{4} - 30 x^{2} ln (x) - 10 x^{2} - 10 x + \frac{25}{x}

= 10 x^{4} - 30 x^{2} ln (x) - 10 x^{2} - 10 x + \frac{25}{x}

\frac{d}{d x} (\frac{x}{csc ( x )})

f^{^{'}} (x) = \frac{e ^{x}}{x}

\frac{\partial}{\partial y} (3 z e^{x y})

\int \frac{2 x + 32}{x ^{2} + 8 x + 7} d x

y = \frac{1}{2} x^{3}

What is the derivative of (2x^3-5(x^2-5ln(x))) ?
The derivative of (2x^3-5(x^2-5ln(x))) is 10x^4-30x^2ln(x)-10x^2-10x+(25)/x
What is the first derivative of (2x^3-5(x^2-5ln(x))) ?
The first derivative of (2x^3-5(x^2-5ln(x))) is 10x^4-30x^2ln(x)-10x^2-10x+(25)/x