Question - Derivative of a Function Involving Square Roots and Chains
Solution:
The function is $$ f(x) = x \sqrt{1 + \sqrt{2x + 6}} $$. To find $$ f'(x) $$, we'll need to use the product rule and the chain rule.The product rule states that if you have a function $$ u(x) \cdot v(x) $$, then the derivative of this function with respect to $$ x $$ is $$ u'(x) \cdot v(x) + u(x) \cdot v'(x) $$.The chain rule states that the derivative of a composite function $$ u(v(x)) $$ is $$ u'(v(x)) \cdot v'(x) $$.Let $$ u(x) = x $$ and $$ v(x) = \sqrt{1 + \sqrt{2x + 6}} $$.Then $$ u'(x) = 1 $$ since the derivative of $$ x $$ with respect to $$ x $$ is 1.For $$ v(x) $$, let $$ h(x) = 1 + \sqrt{2x + 6} $$, so $$ v(x) = \sqrt{h(x)} $$.First, find the derivative of $$ h(x) $$:$$ h'(x) = 0 + \frac{1}{2} (2x + 6)^{-\frac{1}{2}} \cdot (2) = \frac{1}{\sqrt{2x + 6}} $$.Next, we use the chain rule to find the derivative of $$ v(x) $$:$$ v'(x) = \frac{1}{2} h(x)^{-\frac{1}{2}} \cdot h'(x) = \frac{1}{2\sqrt{1 + \sqrt{2x + 6}}} \cdot \frac{1}{\sqrt{2x + 6}} $$.Now we can find $$ f'(x) $$ using the product rule:$$ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) $$$$ f'(x) = 1 \cdot \sqrt{1 + \sqrt{2x + 6}} + x \cdot \left(\frac{1}{2\sqrt{1 + \sqrt{2x + 6}}} \cdot \frac{1}{\sqrt{2x + 6}} \right) $$$$ f'(x) = \sqrt{1 + \sqrt{2x + 6}} + \frac{x}{2(1 + \sqrt{2x + 6})\sqrt{2x + 6}} $$.This is the derivative of the given function. Be sure to simplify where possible to get the final expression for $$ f'(x) $$.
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