What is the derivative of 7z+6z^{1/2}-8 ?

The derivative of 7z+6z^{1/2}-8 is 3/(z^{1/2)}+7

What is the first derivative of 7z+6z^{1/2}-8 ?

The first derivative of 7z+6z^{1/2}-8 is 3/(z^{1/2)}+7

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derivative of

7 z + 6 z^{\frac{1}{2}} - 8

\frac{3}{z ^{\frac{1}{2}}} + 7

Solution steps

\frac{d}{d z} (7 z + 6 z^{\frac{1}{2}} - 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d z} (7 z) + \frac{d}{d z} (6 z^{\frac{1}{2}}) - \frac{d}{d z} (8)

\frac{d}{d z} (7 z) = 7

\frac{d}{d z} (6 z^{\frac{1}{2}}) = \frac{3}{z ^{\frac{1}{2}}}

\frac{d}{d z} (8) = 0

= 7 + \frac{3}{z ^{\frac{1}{2}}} - 0

Simplify

= \frac{3}{z ^{\frac{1}{2}}} + 7

\frac{d}{d x} ((3 cos (x)) (ln (1 + x)))

\frac{d}{d x} (ln ∣ tan (x) + sec (x) ∣ + C)

(9 x^{2} + 6)^{6}

\int \frac{7 sec ^{2} ( 2 x )}{3 - 5 tan ( 2 x )} d x

\frac{d}{d x} (\frac{2}{1 - 7 x})

What is the derivative of 7z+6z^{1/2}-8 ?
The derivative of 7z+6z^{1/2}-8 is 3/(z^{1/2)}+7
What is the first derivative of 7z+6z^{1/2}-8 ?
The first derivative of 7z+6z^{1/2}-8 is 3/(z^{1/2)}+7