# Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 4

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Check by differentiation $\displaystyle\frac{x}{\sqrt{a+bx}} dx = \frac{2}{3b^2} (bx - 2a) \sqrt{a + bx} + C$

We take the derivative of the expression to the right integral

\begin{aligned} f(x) &=\frac{2}{3b^2} (bx - 2a) \sqrt{a+bx} + C \\ \\ f(x) &= \left( \frac{3bx}{3b^2} - \frac{4a}{3b^2} \right) \sqrt{a+bx} + C\\ \\ f(x) &= \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) (a + bx)^{\frac{1}{2}} + C \\ \\ f'(x) &= \left[ \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) \frac{d}{dx} (a + bx)^{\frac{1}{2}} + (a + bx)^{\frac{1}{2}} \frac{d}{dx} \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) \right] + \frac{d}{dx} C\\ \\ f'(x) &= \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) \left( \frac{1}{2} \right) (a + bx)^{\frac{-1}{2}} \frac{d}{dx} ( a + bx) + (a + bx)^{\frac{1}{2}} \left( \frac{2}{3b} - 0 \right) + 0 \\ \\ f'(x) &= \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) \left( \frac{1}{2} \right) (a + bx)^{\frac{-1}{2}} (b)+ (a+bx)^{\frac{1}{2}} \left( \frac{2}{3b} \right)\\ \\ f'(x) &= \frac{\left(\frac{2x}{3b} - \frac{4a}{3b^2} \right)\left( \frac{b}{2} \right)}{(a+bx)^{\frac{1}{2}}} + (a + bx)^{\frac{1}{2}} \left( \frac{2}{3b} \right)\\ \\ f'(x) &= \frac{\frac{2 \cancel{b}x}{6\cancel{b}} - \frac{4ab}{3b^2} }{(a + bx)^{\frac{1}{2}}} + (a + bx)^{\frac{1}{2}} \left( \frac{2}{3b} \right) \\ \\ f'(x) &= \frac{\frac{x}{3} - \cancel{\frac{2a}{3b}} + \cancel{\frac{2a}{3b}} + \frac{2 \cancel{b}x}{3 \cancel{b}} }{(a + bx)^{\frac{1}{2}}}\\ \\ f'(x) &= \frac{\frac{x}{3}+\frac{2x}{3} }{(a+bx)^{\frac{1}{2}}}\\ \\ f'(x) &= \frac{\frac{x+2x}{3}}{(a+bx)^{\frac{1}{2}}}\\ \\ f'(x) &= \frac{\frac{\cancel{3}x}{\cancel{3}}}{(a+bx)^{\frac{1}{2}}}\\ \\ f'(x) &= \frac{x}{(a+bx)^{\frac{1}{2}}} \qquad \text{ or } \qquad f'(x) = \frac{x}{\sqrt{a+bx}} \end{aligned}

The expression to the left is correct