What is the first (3-2^x)^' ?

The first (3-2^x)^' is -ln(2)*2^x

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(3-2^x)^'

(3 - 2^{x})^{^{'}}

- ln (2) \cdot 2^{x}

Solution steps

(3 - 2^{x})^{'}

Apply the Sum/Difference Rule:

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

= 3^{^{'}} - (2^{x})^{^{'}}

3^{'} = 0

(2^{x})^{'} = 2^{x} ln (2)

= 0 - 2^{x} ln (2)

Simplify

= - ln (2) \cdot 2^{x}

\frac{cos ( x ^{3} )}{2}

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6 sin^{5} (x) cos (x)