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

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

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

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

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Popular >

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int cos (8 t) d t

\int \frac{x ^{2} + 2 x - 1}{x ( x + 1 ) ( x - 1 )} d x

y^{^{'}} = 3 x + y, y (0) = 7

x \to 0 + lim (x^{2} csc^{2} (x))

x \frac{d y}{d x} = y - \frac{1}{y}

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