What is the derivative of (2x-y^4-y^3-8) ?

The derivative of (2x-y^4-y^3-8) is 4(2x-y)^3(-(dy)/(dx)+2)-3y^2(dy)/(dx)

What is the first derivative of (2x-y^4-y^3-8) ?

The first derivative of (2x-y^4-y^3-8) is 4(2x-y)^3(-(dy)/(dx)+2)-3y^2(dy)/(dx)

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

4 (2 x - y)^{3} (- \frac{d y}{d x} + 2) - 3 y^{2} \frac{d y}{d x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (8) = 0

= 4 (2 x - y)^{3} (2 - \frac{d y}{d x}) - 3 y^{2} \frac{d y}{d x} - 0

Simplify

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

t^{3} \cdot t^{4}

lo g_{21} (81)

\frac{x ( x + 2 )}{2 ( x + 1 )}

2 (x - 3 π + 2 π^{2})

- 12 x + 27 x

What is the derivative of (2x-y^4-y^3-8) ?
The derivative of (2x-y^4-y^3-8) is 4(2x-y)^3(-(dy)/(dx)+2)-3y^2(dy)/(dx)
What is the first derivative of (2x-y^4-y^3-8) ?
The first derivative of (2x-y^4-y^3-8) is 4(2x-y)^3(-(dy)/(dx)+2)-3y^2(dy)/(dx)