What is the first d/(dT)(7/6 T+17) ?

The first d/(dT)(7/6 T+17) is 7/6

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d/(dT)(7/6 T+17)

\frac{d}{d T} (\frac{7}{6} T + 17)

\frac{7}{6}

Solution steps

\frac{d}{d T} (\frac{7}{6} T + 17)

Simplify

\frac{7}{6} T + 17 : \frac{7 T}{6} + 17

= \frac{d}{d T} (\frac{7 T}{6} + 17)

Apply the Sum/Difference Rule:

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

= \frac{d}{d T} (\frac{7 T}{6}) + \frac{d}{d T} (17)

\frac{d}{d T} (\frac{7 T}{6}) = \frac{7}{6}

\frac{d}{d T} (17) = 0

= \frac{7}{6} + 0

Simplify

= \frac{7}{6}

\frac{d y}{d x} = \frac{y x + y}{y x + x}

x d x - y d y = 0

\frac{\partial}{\partial z} (e^{y})

y = x^{4} - 9 x^{2}, y = x^{2} - 9

\int 6 (t + 1)^{\frac{1}{2}} d t