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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\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}

f (x) = x^{2} cos (x^{2})

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

\int \frac{ln ( x ^{3} )}{x} d x

x \to 2 lim (\frac{2 - ∣ x ∣}{2 + x})

\int_{- \frac{π}{3}}^{0} 3 sec^{3} (x) d 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