What is (\partial)/(\partial b)((h)(2^a(1+b))) ?

The answer to (\partial)/(\partial b)((h)(2^a(1+b))) is 2^ah

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(\partial)/(\partial b)((h)(2^a(1+b)))

\frac{\partial}{\partial b} ((h) (2^{a} (1 + b)))

2^{a} h

Solution steps

\frac{\partial}{\partial b} ((h) (2^{a} (1 + b)))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= h \cdot 2^{a} \frac{\partial}{\partial b} (1 + b)

Apply the Sum/Difference Rule:

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

= h \cdot 2^{a} (\frac{\partial}{\partial b} (1) + \frac{\partial}{\partial b} (b))

\frac{\partial}{\partial b} (1) = 0

\frac{\partial}{\partial b} (b) = 1

= h \cdot 2^{a} (0 + 1)

Simplify

= 2^{a} h

\frac{( x - 3 ) ^{4} ( x + 4 ) ^{3} ( 2 x + 1 ) ^{2} ( 2 x - 1 ) ^{2}}{16}

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

2 e^{- t} + e^{- t} (e^{t} - e^{z})

\frac{( 3 + x ) ^{2}}{( 3 - x ^{2} ) ^{4}}

(3 y^{3} + 4 y x + 1 + 2 x^{2} + 9 y^{2} x)

What is (\partial)/(\partial b)((h)(2^a(1+b))) ?
The answer to (\partial)/(\partial b)((h)(2^a(1+b))) is 2^ah