Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (j)

MATH100 December 2011

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[hide] Question 01 (j)
Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answers. Unless otherwise stated, it is not necessary to simplify your answers in this question. Find f’(x), where $f(x)=x^{(x^{2})}$

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[show] Hint
Can we use the power rule here? Why, or rather, why not?

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[show] 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. In this case we need to use logarithmic differentiation, indeed it is impossible to use the power rule since there is an $x^{2}$ as exponent. We cannot differentiate it in the same way we do for exponential functions since we take the power of x and not e or another constant. Hence we need to use logarithmic differentiation. First set $y=x^{(x^{2})},$ then taking the logarithm on both sides we find $\ln y=\ln(x^{(x^{2})})=x^{2}\ln(x).$ Now keeping in mind that y is a function of x, we can take the derivatives on both sides and get ${\frac {d}{dx}}\ln y={\frac {d}{dx}}(x^{2}\ln(x)).$ The left-hand side is simply ${\frac {d}{dx}}\ln y={\frac {1}{y}}\cdot {\frac {d}{dx}}y={\frac {1}{y}}y',$ while the right-hand side gives, using the product rule, ${\frac {d}{dx}}(x^{2}\ln(x))=2x\ln(x)+x^{2}{\frac {1}{x}}=2x\ln(x)+x.$ Putting everything back together, we get ${\frac {1}{y}}y'=2x\ln(x)+x,$ and thus $\displaystyle y'=y(2x\ln(x)+x).$ Replacing y by $x^{(x^{2})},$ we finally get $y'=x^{(x^{2})}(2x\ln(x)+x).$