If $f(x) = \sqrt{ 4\sin x + 2 }$, then $f'(0) =$?
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Do you know the chain rule? What have you tried? |
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By definition, $$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{\sqrt{4\sin(h)+2}-\sqrt2}{h}.$$ You can "rationalize" the numerator, use the fact that a limit of a product is the product of the limits, use continuity of the square root and sine functions, and use the fact that $\lim_{h\to0}\frac{\sin(h)}{h}=1$ to finish. Of course, you would get the same answer by deriving a general formula for $f'(x)$ using the chain rule and plugging in $x=0$, as Ross hints. |
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$$f'(x)=\frac{4\cos x}{2\sqrt{4\sin x+2}}$$ $$f'(0)=\frac{4\cdot1}{2\sqrt{4\cdot0+2}}=\frac{4}{2\sqrt{2}}=\sqrt{2}$$ what is the difficulty? It does not even involve a $\frac{0}{0}$ like $\sin{x}/x$ does. |
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