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

The derivative of y=(x^2+8)^2-4x is 4x(x^2+8)-4

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

The first derivative of y=(x^2+8)^2-4x is 4x(x^2+8)-4

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derivative of

y = (x^{2} + 8)^{2} - 4 x

4 x (x^{2} + 8) - 4

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} ((x^{2} + 8)^{2}) = 4 x (x^{2} + 8)

\frac{d}{d x} (4 x) = 4

= 4 x (x^{2} + 8) - 4

\frac{\partial}{\partial x} (9 csc (6 x^{4} - 9 x + 5))

\frac{d y}{d x}, e^{\frac{x}{y}} = 6 x - y

\frac{d}{d x} (\frac{8}{( 1 - 2 x ) ^{3}})

\frac{d}{d x} (\frac{x ^{5}}{10})

y = 6 - x^{2}, y = 8 - 3 x

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