What is the derivative of (2016^u)(u+2017) ?

The derivative of (2016^u)(u+2017) is ln(2016)*2016^u(u+2017)+2016^u

What is the first derivative of (2016^u)(u+2017) ?

The first derivative of (2016^u)(u+2017) is ln(2016)*2016^u(u+2017)+2016^u

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derivative of

(201 6^{u}) (u + 2017)

ln (2016) \cdot 201 6^{u} (u + 2017) + 201 6^{u}

Solution steps

\frac{d}{d u} (201 6^{u} (u + 2017))

Apply the Product Rule:

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

f = 201 6^{u}, g = u + 2017

= \frac{d}{d u} (201 6^{u}) (u + 2017) + \frac{d}{d u} (u + 2017) 201 6^{u}

\frac{d}{d u} (201 6^{u}) = 201 6^{u} ln (2016)

\frac{d}{d u} (u + 2017) = 1

= 201 6^{u} ln (2016) (u + 2017) + 1 \cdot 201 6^{u}

Simplify

= ln (2016) \cdot 201 6^{u} (u + 2017) + 201 6^{u}

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

\frac{d}{d t} (\frac{( t ^{3} - 6 t ^{2} + 7 )}{( 2 t ^{2} + 6 t )})

e^{3 x + 2} + 5

(((e^{e})^{e})^{x})^{2}

5 x^{3} + 16 b^{2}

What is the derivative of (2016^u)(u+2017) ?
The derivative of (2016^u)(u+2017) is ln(2016)*2016^u(u+2017)+2016^u
What is the first derivative of (2016^u)(u+2017) ?
The first derivative of (2016^u)(u+2017) is ln(2016)*2016^u(u+2017)+2016^u