# Taking the Derivative of $\frac {\sin(x^2)}{3x}$, which step is wrong?

So I'm trying to take the derivative of $\frac {\sin(x^2)}{3x}$.

Here are my steps:

$$\frac {d}{dx}\left[\frac {\sin(x^2)}{3x}\right]$$ Use the Quotient Rule: $$\frac {3x\frac {d}{dx}[\sin(x^2)]-\sin(x^2)\frac {d}{dx}[3x]}{3x^2}$$

Simplify second part:

$$\frac {3x\frac {d}{dx} [\sin(x^2)]-3\sin(x^2)}{3x^2}$$

Use Chain Rule on first part:

$$\frac {d}{dx} [\sin(x^2)]= 2x(\cos(x^2))$$

Plug it into the numerator:

$$\frac {3x(2x)\cos(x^2)-3\sin(x^2)}{3x^2}$$

Simplify a little / Result:

$$\frac {6x^2\cos(x^2)-3\sin(x^2)}{3x^2}$$

Right answer: $$\frac {2x^2\cos(x^2)-\sin(x^2)}{3x^2}$$

So I messed something up, but I tried to redo the process multiple times and couldn't figure it out. Thanks in advance for any help!

• You have to square the entire denominator. – Adrian Keister Apr 11 '18 at 19:05
• Also, use backslash in your $\LaTeX$ for sin and cosine. – Adrian Keister Apr 11 '18 at 19:06
• Thanks Adrian! :) – user472288 Apr 11 '18 at 19:13
• You can move $3$ in the denominator out of differentiation in the beginning. – farruhota Apr 11 '18 at 19:26

As Adrian said in the comments, when you used quotient rule, the entire denominator should be squared, giving a denominator of $9x^2$ rather than $3x^2$.
This gives $$\frac {6x^2\cos(x^2)-3\sin(x^2)}{9x^2}=\frac {2x^2\cos(x^2)-\sin(x^2)}{3x^2}$$as required.
Edit: Just seen your comment - if you already have parentheses around the $3x$, then your answer is not wrong at all, it just needs to be simplified (as I have done above).
It is $$\frac{\cos(x^2)\cdot 2x\cdot3x-\sin(x^2)\cdot 3}{9x^2}$$ which simplifies to $$\frac{2 x^2 \cos \left(x^2\right)-\sin \left(x^2\right)}{3 x^2}$$