Science:Math Exam Resources/Courses/MATH110/December 2013/Question 04 (c)

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MATH110 December 2013

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Question 04 (c)
Let $f(x)={\sqrt {x}}-1$ and $g(x)=(f\circ f)(x)$. Find the derivative of $g(x)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that $\displaystyle (f\circ f)(x)={\sqrt {{\sqrt {x}}-1}}-1$. This is a composition of functions, namely $\displaystyle {\sqrt {x}}$ and $\displaystyle {\sqrt {x}}-1$ (and then the -1 term is easily differentiated). This can be differentiated using the chain rule.

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Solution

We want to apply the chain rule. Set $u={\sqrt {x}}-1$. ${\begin{aligned}g(u)&={\sqrt {u}}-1&g'(u)&={\frac {1}{2{\sqrt {u}}}}\\u(x)&={\sqrt {x}}-1&u'(x)&={\frac {1}{2{\sqrt {x}}}}\end{aligned}}$ Combining, we get: ${\begin{aligned}g'(x)&={\frac {1}{2{\sqrt {x}}}}\cdot {\frac {1}{2{\sqrt {{\sqrt {x}}-1}}}}\\&={\frac {1}{4{\sqrt {x}}\cdot {\sqrt {{\sqrt {x}}-1}}}}\\&={\frac {1}{4{\sqrt {x\left({\sqrt {x}}-1\right)}}}}\end{aligned}}$

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