What is the first ((7^3*11^5)^{1/4})^' ?

The first ((7^3*11^5)^{1/4})^' is 0

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((7^3*11^5)^{1/4})^'

((7^{3} \cdot 1 1^{5})^{\frac{1}{4}})^{^{'}}

0

Solution steps

((7^{3} \cdot 1 1^{5})^{\frac{1}{4}})^{'}

(7^{3} \cdot 1 1^{5})^{\frac{1}{4}} = 1 1^{\frac{5}{4}} \cdot 7^{\frac{3}{4}}

= (1 1^{\frac{5}{4}} \cdot 7^{\frac{3}{4}})^{^{'}}

Derivative of a constant:

\frac{d}{d x} (a) = 0

= 0

e^{- 2 t} u (t + 1)

3 (\frac{245}{360} \cdot π (3)^{2})

6 (1.3)^{- 1}

4.2 \cdot 1 0^{- 3} \frac{\partial}{\partial x} (34)