What is the derivative of 8xy-(1/4 (x+y)^4) ?

The derivative of 8xy-(1/4 (x+y)^4) is 8(y+x(dy)/(dx))-(1+(dy)/(dx))(x+y)^3

What is the first derivative of 8xy-(1/4 (x+y)^4) ?

The first derivative of 8xy-(1/4 (x+y)^4) is 8(y+x(dy)/(dx))-(1+(dy)/(dx))(x+y)^3

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\frac{d}{d x} (8 x y - (\frac{1}{4}) (x + y)^{4})

8 (y + x \frac{d y}{d x}) - (1 + \frac{d y}{d x}) (x + y)^{3}

Solution steps

\frac{d}{d x} (8 x y - (\frac{1}{4}) (x + y)^{4})

Simplify

8 x y - (\frac{1}{4}) (x + y)^{4} : 8 x y - \frac{( x + y ) ^{4}}{4}

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (8 x y) = 8 (y + x \frac{d y}{d x})

\frac{d}{d x} (\frac{( x + y ) ^{4}}{4}) = (1 + \frac{d y}{d x}) (x + y)^{3}

= 8 (y + x \frac{d y}{d x}) - (1 + \frac{d y}{d x}) (x + y)^{3}

x (1 + y^{2}) d x = y (1 + x^{2}) d y

(- 2, - 7), (3, 5)

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

y^{^{''}} - 3 y^{^{'}} = x^{3} - 6

\frac{d}{d x} (15 cos (3 x))

What is the derivative of 8xy-(1/4 (x+y)^4) ?
The derivative of 8xy-(1/4 (x+y)^4) is 8(y+x(dy)/(dx))-(1+(dy)/(dx))(x+y)^3
What is the first derivative of 8xy-(1/4 (x+y)^4) ?
The first derivative of 8xy-(1/4 (x+y)^4) is 8(y+x(dy)/(dx))-(1+(dy)/(dx))(x+y)^3