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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int \frac{2 t}{1 - t ^{2}} d t

n = 1 \sum \infty n \cdot 2^{- n}

- 5 t^{2} y^{^{''}} - 2 t y^{^{'}} + 14 y = 0

f (x) = π x

x^{2} e^{x} - 2 x e^{x} + 5 e^{x}

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