# Question on derivatives of trig functions

• March 20th 2010, 02:41 AM
nozpoz
Question on derivatives of trig functions
I have a maths B (Grade 11 & 12 Advanced) exam on monday and im studying up for my whole weekend. D=
Anyway one question that I need help for is:

y = tan^2 3x

The answer I got in the end after using the quotient rule was

dy/dx = 6cos3x * cos^2 3x + sin^2 3x * 6sin 3x / cos^3 3x

The denominator is correct but the answer says that the numerator is 6sin3x

Can you please help me using tanx = sinx / cosx not sec^2 x

Thanks to anyone who can help me =)
• March 20th 2010, 04:15 AM
Soroban
Hello, nozpoz!

You dropped some exponents . . .

Quote:

Differentiate with the Quotient Rule: . $y \:=\:\tan^23x$

We have: . $y \;=\;\frac{\sin^23x}{\cos^23x}$

$\text{Then: }\;\frac{dy}{dx} \;=\;\frac{\cos^23x\cdot \bigg[2\cdot\sin3x\cdot\cos3x\cdot 3\bigg] - \sin^23x\cdot\bigg[2\cdot\cos3x(-\sin3x)\cdot 3\bigg]}{\cos^43x}$

. . . . . . . $=\;\frac{6\sin3x\cos^33x + 6\sin^33x\cos3x}{\cos^43x}$

. . . . . . . $=\;\frac{6\sin3x\cos3x\cdot\overbrace{(\cos^23x + \sin^23x)}^{\text{This is 1}}}{\cos^43x}$

. . . . . . . $=\;\frac{6\sin3x}{\cos^33x}$

which is equal to: . $6\sec^23x\tan3x$