What is the derivative of G/4+sqrt(W)G ?

The derivative of G/4+sqrt(W)G is 1/4+sqrt(W)

What is the first derivative of G/4+sqrt(W)G ?

The first derivative of G/4+sqrt(W)G is 1/4+sqrt(W)

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derivative of

\frac{G}{4} + W G

\frac{1}{4} + W

Solution steps

\frac{d}{d G} (\frac{G}{4} + W G)

Treat

W

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d G} (\frac{G}{4}) + \frac{d}{d G} (W G)

\frac{d}{d G} (\frac{G}{4}) = \frac{1}{4}

\frac{d}{d G} (W G) = W

= \frac{1}{4} + W

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 3 y^{3}))

sin (10 t)

x \to 0 lim (\frac{8}{x} - \frac{8}{∣ x ∣})

\frac{d}{d x} (360 x^{2} - 480 x + 120)

\frac{d}{d x} (y^{2} (x^{2} + 1))

What is the derivative of G/4+sqrt(W)G ?
The derivative of G/4+sqrt(W)G is 1/4+sqrt(W)
What is the first derivative of G/4+sqrt(W)G ?
The first derivative of G/4+sqrt(W)G is 1/4+sqrt(W)