Find the derivative. $$ \frac{d}{d x} \int_{1}^{\sqrt{x}} 18 t^{5} d t $$

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Find the derivative with respect to $$ x $$ for $$ \int_{1}^{\sqrt{x}} 18t^5 dt $$.
Evaluate the derivative by following steps:
- step0: Evaluate the derivative:
  $$\frac{d}{dx}\left(\int_{1}^{\sqrt{x}} 18t^{5} dtight)$$
- step1: Use the substitution:
  $$\frac{d}{dx}\left(\int_{1}^{v} 18t^{5} dtight)$$
- step2: Use differentiation rules:
  $$\frac{d}{dv}\left(\int_{1}^{v} 18t^{5} dtight)	imes \frac{dv}{dx}$$
- step3: Apply the theorem:
  $$18v^{5}	imes \frac{dv}{dx}$$
- step4: Rewrite the derivative:
  $$18v^{5}	imes \frac{d}{dx}\left(vight)$$
- step5: Substitute back:
  $$18\left(\sqrt{x}ight)^{5}	imes \frac{d}{dx}\left(\sqrt{x}ight)$$
- step6: Find the derivative:
  $$18\left(\sqrt{x}ight)^{5}	imes \frac{1}{2\sqrt{x}}$$
- step7: Calculate:
  $$18x^{2}\sqrt{x}	imes \frac{1}{2\sqrt{x}}$$
- step8: Reduce the fraction:
  $$9x^{2}	imes 1$$
- step9: Multiply the terms:
  $$9x^{2}$$
The derivative of $$ \frac{d}{dx} \int_{1}^{\sqrt{x}} 18t^{5} dt $$ with respect to $$ x $$ is $$ 9x^{2} $$.
The derivative is $$ 9x^{2} $$.

Find the derivative.

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