Science:Math Exam Resources/Courses/MATH110/December 2011/Question 11

MATH110 December 2011

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Question 11
Find a piecewise algebraic expression for the nth derivative of $f(x)=x\ln x$. Note: This question was for bonus marks, so it may be more challenging.

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
Write out the first few derivatives of $f$. What pattern do you notice?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We compute the first few derivatives of $f(x)=x\ln(x)$ and look for a pattern. ${\begin{array}{ccl}f'(x)&=&\ln(x)+1\\f''(x)&=&1/x\\f'''(x)&=&-1/x^{2}\\f^{(4)}(x)&=&2/x^{3}\\f^{(5)}(x)&=&-6/x^{3}\\\end{array}}$ After the second derivative, we notice that the derivative is a fraction, with a constant in the numerator and a power of x in the denominator. In fact, we can write this pattern as $f^{(n)}(x)={\frac {(n-2)!(-1)^{n}}{x^{n-1}}}$ for all n ≥ 2. The only derivative that is different is the first one. Thus, our piecewise function describing the derivative is: <math> f^{(n)}(x) = \begin{cases} \ln(x) & n = 1 \\ \frac{ (n-2)! (-1)^n }{x^{n-1}} & n > 1 \end{cases}

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