Question

f(x) = sqrt(x^3+4)
f '(x) = ?

Answer 1

$</p><p> \sf \quad f(x) = \sqrt{ {x}^{3} + 4} \\ \\ \\ \sf \rightarrow \quad f'(x) = \frac{d}{dx} \left( \sqrt{ {x}^{3} + 4} \right) \\ \\ \sf \rightarrow \quad f'(x) = \frac{d( {x}^{3} + 4 )^{ \frac{1}{2} } }{dx} \\ \\ \sf \rightarrow \quad f'(x) = \frac{d( {x}^{3} + 4 )^{ \frac{1}{2} } }{d( {x}^{3} + 4)} \: \cdot \: \frac{d( {x}^{3} + 4)}{dx} \\ \\ \sf \rightarrow \quad f'(x) = \frac{1}{2} ( {x}^{3} + 4)^{ - \frac{1}{2} } \: \cdot \: \left( \frac{d( {x}^{3}) }{dx} + \frac{d(4)}{dx} \right) \\ \\ \sf \rightarrow \quad f'(x) = \frac{1}{2 \sqrt{ {x}^{3} + 4} } \: \cdot \: (3 {x}^{2} + 0) \\ \\ \sf \rightarrow \quad f'(x) = \frac{3 {x}^{2} }{2 \sqrt{ {x}^{3} + 4} } </p><p>$