What is expand (dy^3)/(d^3x)(y=cos(x)-ln(2x)) ?

The solution to expand (dy^3)/(d^3x)(y=cos(x)-ln(2x)) is (y^4)/(d^2x)=cos(x)-(ln(2x)y^3)/(d^2x)

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expand (dy^3)/(d^3x)(y=cos(x)-ln(2x))

expand

\frac{d y ^{3}}{d ^{3} x} (y = cos (x) - ln (2 x))

\frac{y ^{4}}{d ^{2} x} = cos (x) - \frac{ln ( 2 x ) y ^{3}}{d ^{2} x}

Solution steps

\frac{d y ^{3}}{d ^{3} x} (y = cos (x) - ln (2 x))

= \frac{d y ^{3}}{d ^{3} x} (= cos (x) y - ln (2 x))

Apply the distributive law:

a (b - c) = ab - a c

a = \frac{d y ^{3}}{d ^{3} x}, b = y = cos (x), c = ln (2 x)

\frac{d y ^{3}}{d ^{3} x} y = cos (x) - \frac{d y ^{3}}{d ^{3} x} ln (2 x)

Simplify

\frac{d y ^{3}}{d ^{3} x} y = cos (x) - \frac{d y ^{3}}{d ^{3} x} ln (2 x) : \frac{y ^{4}}{d ^{2} x} = cos (x) - \frac{ln ( 2 x ) y ^{3}}{d ^{2} x}

\frac{y ^{4}}{d ^{2} x} = cos (x) - \frac{ln ( 2 x ) y ^{3}}{d ^{2} x}

lo g_{9} (x + 3) (x - 5)

(3 X + 2 Y - 16 = 0,) (5 X + 2 Y - (- 18) = 0)

(ln (t))^{2}

(2 x + 2 y y^{'} (x) = 0) \frac{1}{2}

(1 + t)^{2} (1 - t)^{3}

What is expand (dy^3)/(d^3x)(y=cos(x)-ln(2x)) ?
The solution to expand (dy^3)/(d^3x)(y=cos(x)-ln(2x)) is (y^4)/(d^2x)=cos(x)-(ln(2x)y^3)/(d^2x)