What is the derivative of y=ln(sqrt(5x)*(2x^3-4x)^5) ?

The derivative of y=ln(sqrt(5x)*(2x^3-4x)^5) is (31x^2-22)/(2x(x^2-2))

What is the first derivative of y=ln(sqrt(5x)*(2x^3-4x)^5) ?

The first derivative of y=ln(sqrt(5x)*(2x^3-4x)^5) is (31x^2-22)/(2x(x^2-2))

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derivative of

y = ln (5 x \cdot (2 x^{3} - 4 x)^{5})

\frac{31 x ^{2} - 22}{2 x ( x ^{2} - 2 )}

Solution steps

\frac{d}{d x} (ln (5 x (2 x^{3} - 4 x)^{5}))

Simplify

ln (5 x (2 x^{3} - 4 x)^{5}) : ln (x (2 x^{3} - 4 x)^{5}) + \frac{1}{2} ln (5)

= \frac{d}{d x} (ln (x (2 x^{3} - 4 x)^{5}) + \frac{1}{2} ln (5))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (ln (x (2 x^{3} - 4 x)^{5})) + \frac{d}{d x} (\frac{1}{2} ln (5))

\frac{d}{d x} (ln (x (2 x^{3} - 4 x)^{5})) = \frac{31 x ^{2} - 22}{2 x ( x ^{2} - 2 )}

\frac{d}{d x} (\frac{1}{2} ln (5)) = 0

= \frac{31 x ^{2} - 22}{2 x ( x ^{2} - 2 )} + 0

Simplify

= \frac{31 x ^{2} - 22}{2 x ( x ^{2} - 2 )}

\int 1 + e^{x} sin (y) d x

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

y^{^{'}} = 3 x

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

x^{3} + e^{x}

What is the derivative of y=ln(sqrt(5x)*(2x^3-4x)^5) ?
The derivative of y=ln(sqrt(5x)*(2x^3-4x)^5) is (31x^2-22)/(2x(x^2-2))
What is the first derivative of y=ln(sqrt(5x)*(2x^3-4x)^5) ?
The first derivative of y=ln(sqrt(5x)*(2x^3-4x)^5) is (31x^2-22)/(2x(x^2-2))