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I'm learning Calculus and I don't know how the solution is wrong, I'm trying the below: $$f(t) = \frac{1}{\sqrt{1 + 5t^{2}}} \ \rightarrow \ (1+5t^2)^\frac{-1}{2} $$ I know I should use a combination of the power and chain rule: $$f'(t) = -\frac{1}{2}(1+5t^2)^\frac{-3}{2} \times 10t$$ This is where I'm stuck

Here is the correct solution, and I don't understand how they got it: $$f'(t) = -\frac{5t}{ (5t^{2} + 1)^\frac{3}{2} } \ $$

UPDATE: They simplified, had a moment there thanks!

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    $\begingroup$ That's it. You're done! $\endgroup$ – Sean Roberson Jun 10 '18 at 17:14
  • $\begingroup$ Nvm just saw my trivial mistake thanks! $\endgroup$ – craz1001 Jun 10 '18 at 17:20
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    $\begingroup$ Your solution is equal to the correct solution and therefore it is correct. $\endgroup$ – José Carlos Santos Jun 10 '18 at 17:20
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    $\begingroup$ @craz1001 Minor point but you've got a function of $x$ on the left-side but a function of $t$ on the right-side. $\endgroup$ – Jam Jun 10 '18 at 17:29
  • $\begingroup$ Glad you found your mistake. Now please delete the question, since leaving it here will probably not help anyone else in the future. $\endgroup$ – Ethan Bolker Jun 10 '18 at 17:41
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Also, you could use the quotient rule: the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator all divided by the denominator squared.

$$f'(t) = \frac{0 - 1\cdot(-\frac{1}{2}(1+5t^2)^{-\frac{1}{2}}\cdot 10t}{1+5t^2} = \frac{-5t}{(1+5t^2)^{\frac{3}{2}}}$$

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