What is the derivative of y=f(u)=u(y,x)^3-4yu(y,x)=g(x)=x^3+3 ?

The derivative of y=f(u)=u(y,x)^3-4yu(y,x)=g(x)=x^3+3 is 3x^2

What is the first derivative of y=f(u)=u(y,x)^3-4yu(y,x)=g(x)=x^3+3 ?

The first derivative of y=f(u)=u(y,x)^3-4yu(y,x)=g(x)=x^3+3 is 3x^2

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derivative of

y = f (u) = u (y, x) (y, x)^{3} - 4 y u (y, x) = g (x) = x^{3} + 3

3 x^{2}

Solution steps

\frac{d}{d x} (x^{3} + 3)

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (3) = 0

= 3 x^{2} + 0

Simplify

= 3 x^{2}

f (x) = x^{2} - 2, at x = - 5, 23

x \to 3 π lim (e^{s i n (x)})

x \to 3 lim (\frac{∣ x - 3 ∣}{3 - x})

\int (1 - 2 x) e^{- 2 x} d x

What is the derivative of y=f(u)=u(y,x)^3-4yu(y,x)=g(x)=x^3+3 ?
The derivative of y=f(u)=u(y,x)^3-4yu(y,x)=g(x)=x^3+3 is 3x^2
What is the first derivative of y=f(u)=u(y,x)^3-4yu(y,x)=g(x)=x^3+3 ?
The first derivative of y=f(u)=u(y,x)^3-4yu(y,x)=g(x)=x^3+3 is 3x^2