What is the derivative of ax^4-2ax^3+2bx^2*ln(x/2) ?

The derivative of ax^4-2ax^3+2bx^2*ln(x/2) is 4ax^3-6ax^2+2b(2xln(x/2)+x)

What is the first derivative of ax^4-2ax^3+2bx^2*ln(x/2) ?

The first derivative of ax^4-2ax^3+2bx^2*ln(x/2) is 4ax^3-6ax^2+2b(2xln(x/2)+x)

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\frac{d}{d x} (a x^{4} - 2 a x^{3} + 2 b x^{2} \cdot ln (\frac{x}{2}))

4 a x^{3} - 6 a x^{2} + 2 b (2 x ln (\frac{x}{2}) + x)

Solution steps

\frac{d}{d x} (a x^{4} - 2 a x^{3} + 2 b x^{2} ln (\frac{x}{2}))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (a x^{4}) - \frac{d}{d x} (2 a x^{3}) + \frac{d}{d x} (2 b x^{2} ln (\frac{x}{2}))

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

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

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

= 4 a x^{3} - 6 a x^{2} + 2 b (2 x ln (\frac{x}{2}) + x)

x + 2

x = - y^{2} + 4 y, [1, 4]

x \to 0 lim (sin^{2} (\frac{π}{x}))

\frac{\partial}{\partial x} ((9 x + 2 y)^{10})

f (x) = e^{2 x} + 1

What is the derivative of ax^4-2ax^3+2bx^2*ln(x/2) ?
The derivative of ax^4-2ax^3+2bx^2*ln(x/2) is 4ax^3-6ax^2+2b(2xln(x/2)+x)
What is the first derivative of ax^4-2ax^3+2bx^2*ln(x/2) ?
The first derivative of ax^4-2ax^3+2bx^2*ln(x/2) is 4ax^3-6ax^2+2b(2xln(x/2)+x)