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Question

Find the value of h′$$(e)h^{\prime}$$\left(e\right) where h(x)=(3x)13xh\left(x\right)=\left(3x\$$right)^{\frac{1}$${3x}}, and ee is the base of the natural logarithm.

Accepted Answer

h′(e)=−(3e)13eln⁡$$33e2h^{\prime}$$\left(e\right)=-\frac{\left(3e\$$right)^{\frac{1}$${3e}}\ln3}{$$3e^2$$}

Answer

h′(e)=(3e)13eln⁡$$33e2h^{\prime}$$\left(e\right)=\frac{\left(3e\$$right)^{\frac{1}$${3e}}\ln3}{$$3e^2$$}

Answer

h′(e)=−(3e)13eln⁡$$3e2h^{\prime}$$\left(e\right)=-\frac{\left(3e\$$right)^{\frac{1}$${3e}}\ln3}{$$e^2$$}

Answer

h′(e)=(3e)−13eln⁡$$3e2h^{\prime}$$\left(e\right)=\frac{\left(3e\$$right)^{-\frac{1}$${3e}}\ln3}{$$e^2$$}

Find the value of h′(e)h^{\(\prime\)}\(\left\)(e\(\right\)) where h(x)=(3x)13xh\(\left\)(x\(\right\))=\(\left\)(3x\(\right\))^{\(\frac{1}{3x}\)}, and ee is the base of the natural logarithm.

Find the value of $h^{'} (e)$ where $h (x) = {(3 x)}^{\frac{1}{3 x}}$ , and $e$ is the base of the natural logarithm.