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

The derivative of ax^3ln(x/4+bx^2(x-4)) is a(3x^2ln(x/4)+x^2)+b(3x^2-8x)

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

The first derivative of ax^3ln(x/4+bx^2(x-4)) is a(3x^2ln(x/4)+x^2)+b(3x^2-8x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

a (3 x^{2} ln (\frac{x}{4}) + x^{2}) + b (3 x^{2} - 8 x)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (b x^{2} (x - 4)) = b (3 x^{2} - 8 x)

= a (3 x^{2} ln (\frac{x}{4}) + x^{2}) + b (3 x^{2} - 8 x)

ln (3 x^{5})

\frac{d}{d x} (\frac{e ^{x}}{4 - 2 x})

\frac{d y}{d t} = \frac{( y + 3 )}{( 1 - y )}

\int 2 csc (x) cot (x) d x

\int 11 θ tan^{2} (θ) d θ

What is the derivative of ax^3ln(x/4+bx^2(x-4)) ?
The derivative of ax^3ln(x/4+bx^2(x-4)) is a(3x^2ln(x/4)+x^2)+b(3x^2-8x)
What is the first derivative of ax^3ln(x/4+bx^2(x-4)) ?
The first derivative of ax^3ln(x/4+bx^2(x-4)) is a(3x^2ln(x/4)+x^2)+b(3x^2-8x)