\require{AMSmath} Zonder kettingregel $ \eqalign{ & f(x) = \sqrt {2x^2 + x} \cr & \sqrt {2x^2 + x} \cdot \sqrt {2x^2 + x} = 2x^2 + x \cr & \left[ {\sqrt {2x^2 + x} } \right]^| \cdot \sqrt {2x^2 + x} + \sqrt {2x^2 + x} \left[ {\sqrt {2x^2 + x} } \right]^| = 4x + 1 \cr & 2 \cdot \left[ {\sqrt {2x^2 + x} } \right]^| \cdot \sqrt {2x^2 + x} = 4x + 1 \cr & \left[ {\sqrt {2x^2 + x} } \right]^| = \frac{{4x + 1}} {{2\sqrt {2x^2 + x} }} \cr & f'(x) = = \frac{{4x + 1}} {{2\sqrt {2x^2 + x} }} \cr} $ Met de productregel Met machten Standaardafgeleide ©2004-2024 WisFaq
\require{AMSmath}
$ \eqalign{ & f(x) = \sqrt {2x^2 + x} \cr & \sqrt {2x^2 + x} \cdot \sqrt {2x^2 + x} = 2x^2 + x \cr & \left[ {\sqrt {2x^2 + x} } \right]^| \cdot \sqrt {2x^2 + x} + \sqrt {2x^2 + x} \left[ {\sqrt {2x^2 + x} } \right]^| = 4x + 1 \cr & 2 \cdot \left[ {\sqrt {2x^2 + x} } \right]^| \cdot \sqrt {2x^2 + x} = 4x + 1 \cr & \left[ {\sqrt {2x^2 + x} } \right]^| = \frac{{4x + 1}} {{2\sqrt {2x^2 + x} }} \cr & f'(x) = = \frac{{4x + 1}} {{2\sqrt {2x^2 + x} }} \cr} $
Met de productregel Met machten Standaardafgeleide
©2004-2024 WisFaq