1. **State the problem:** We are given the function $$f(x) = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$$ where $$a > 0$$ and its derivative $$f'(x) = \frac{1}{x^2 + 49}$$. We need to prove that $$a = 7$$. 2. **Recall the derivative formula:** The derivative of $$\tan^{-1}(u)$$ with respect to $$x$$ is $$\frac{1}{1+u^2} \cdot \frac{du}{dx}$$. 3. **Apply the chain rule to $$f(x)$$:** $$$ f'(x) = \frac{1}{a} \cdot \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{d}{dx}\left(\frac{x}{a}\right) = \frac{1}{a} \cdot \frac{1}{1 + \frac{x^2}{a^2}} \cdot \frac{1}{a} = \frac{1}{a^2} \cdot \frac{1}{1 + \frac{x^2}{a^2}} = \frac{1}{a^2} \cdot \frac{1}{\frac{a^2 + x^2}{a^2}} = \frac{1}{a^2} \cdot \frac{a^2}{a^2 + x^2} = \frac{1}{a^2 + x^2} $$$ 4. **Compare with given derivative:** We have $$$ f'(x) = \frac{1}{a^2 + x^2} \quad \text{and} \quad f'(x) = \frac{1}{x^2 + 49} $$$ 5. **Equate denominators:** Since these are equal for all $$x$$, $$$ a^2 + x^2 = x^2 + 49 $$$ 6. **Simplify:** Cancel $$x^2$$ on both sides: $$$ \cancel{a^2} + \cancel{x^2} = \cancel{x^2} + 49 \implies a^2 = 49 $$$ 7. **Solve for $$a$$:** Since $$a > 0$$, $$$ a = \sqrt{49} = 7 $$$ **Final answer:** $$a = 7$$.