What is the derivative of f(t)=t^{1/2}(13t+2) ?

The derivative of f(t)=t^{1/2}(13t+2) is (39t+2)/(2t^{1/2)}

What is the first derivative of f(t)=t^{1/2}(13t+2) ?

The first derivative of f(t)=t^{1/2}(13t+2) is (39t+2)/(2t^{1/2)}

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derivative of

f (t) = t^{\frac{1}{2}} (13 t + 2)

\frac{39 t + 2}{2 t ^{\frac{1}{2}}}

Solution steps

\frac{d}{d t} (t^{\frac{1}{2}} (13 t + 2))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{\frac{1}{2}}, g = 13 t + 2

= \frac{d}{d t} (t^{\frac{1}{2}}) (13 t + 2) + \frac{d}{d t} (13 t + 2) t^{\frac{1}{2}}

\frac{d}{d t} (t^{\frac{1}{2}}) = \frac{1}{2 t ^{\frac{1}{2}}}

\frac{d}{d t} (13 t + 2) = 13

= \frac{1}{2 t ^{\frac{1}{2}}} (13 t + 2) + 13 t^{\frac{1}{2}}

Simplify

\frac{1}{2 t ^{\frac{1}{2}}} (13 t + 2) + 13 t^{\frac{1}{2}} : \frac{39 t + 2}{2 t ^{\frac{1}{2}}}

= \frac{39 t + 2}{2 t ^{\frac{1}{2}}}

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What is the derivative of f(t)=t^{1/2}(13t+2) ?
The derivative of f(t)=t^{1/2}(13t+2) is (39t+2)/(2t^{1/2)}
What is the first derivative of f(t)=t^{1/2}(13t+2) ?
The first derivative of f(t)=t^{1/2}(13t+2) is (39t+2)/(2t^{1/2)}