0
$\begingroup$

So, there's this function I'm having trouble finding the derivative of:

$y = \frac{-5}{\cos \sqrt{{3x+2x^3}}}$

I checked the back of the textbook, which had the answer as $y' = \frac{-15(1+2x^2)\sin \sqrt{{3x+2x^3}}}{2 \sqrt{3x+2x^3}(\cos \sqrt{{3x+2x^3}})^2}$

However, when I calculated the derivative on my graphing calculator, the answer came out as $y' = -\frac{2.5\sin \sqrt{{3x+2x^3}} (3+6x^2)}{(\cos \sqrt{{3x+2x^3}})^2 \sqrt{3x+2x^3}}$

I have no idea how to reach either of these solutions. Anyone know what's going on here?

  • 0
    Do you know the Quotient Rule for Derivatives? – 2017-02-16

1 Answers 1

0

Well, you have also got the correct answer barring some simplifications. We know that $2.5 =\frac {5}{2} $ and that $(3+6x^2) =3 (1+2x^2) $.

Thus, we get, $$-\frac {2.5\sin \sqrt {3x+2x^3}(3+6x^2)}{(\cos \sqrt {3x+2x^3})^2 \sqrt {3x+2x^3}} $$ $$=-\frac {5}{2} \times \frac {3 (1+2x^2)\sin \sqrt {3x+2x^3}}{(\cos \sqrt {3x+2x^3})^2\sqrt {3x+2x^3}} $$ $$=- \frac {15 (1+2x^2)\sin \sqrt {3x+2x^3}}{2(\cos \sqrt {3x+2x^3})^2\sqrt {3x+2x^3}}$$ which is the same as the books answer. Hope it helps.

  • 0
    Brilliant, that makes perfect sense. Thank you. – 2017-02-16