Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

There seems to be something off with one of my problem solutions. Could somebody find the mistake?

$$x=\sqrt{t^3-1} \ \text{,} \ y=\ln t$$ $$y'(t) = \frac{1}{t}$$ $$x'(t)=\frac{1}{2\sqrt{t^3-1}} \cdot 3t^2 = \frac{3t^2}{2\sqrt{t^3-1}} $$ $$y'_x= \frac{\frac{1}{t}}{\frac{1}{2\sqrt{t^3-1}} \cdot 3t^2} = \frac{2\sqrt{t^3-1}}{3t^3}$$ $$y''_{xx} =\frac{\left(\frac{2\sqrt{t^3-1}}{3t^3}\right)'}{\frac{3t^2}{2\sqrt{t^3-1}}} = \frac{\frac{\frac{1}{\sqrt{t^3-1}}\cdot 3t^2 *3t^3 - 2\sqrt{t^3-1} *9t^2}{9t^6}}{\frac{3t^2}{2\sqrt{t^3-1}}}=\frac{12t^5-36t^2(t^3-1)}{27t^8}=\frac{4t^3-12t^3+12}{9t^6}$$

share|improve this question
is $y'_x = \frac{dy}{dx}$? –  user88595 Oct 12 '13 at 14:08
yes, you are correct. –  vilbur Oct 12 '13 at 14:10
From what I see it seems correct. What makes you think it's wrong? You could check by letting $y = \ln{\big((x^2 + 1)^{1/3}\big)}$ and then substituting back $x = \sqrt{t^3 - 1}$ assuming my inversion is correct. –  user88595 Oct 12 '13 at 14:19
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.