Derivatives Involving Parentheses and Trig Functions

Question

$y=(\cos3)x^{2}$

If we were to take the first derivative of the equation, wouldn't we apply the product rule so that:

$y^{'}=f(x)g^{'}(x)+g(x)f^{'}(x)$, where
$f(x)=\cos3 $ , and
$g(x)=x^2$

Then,

$y^{'}=(\cos3)2x + x^{2}(-\sin 3)$
$=(\cos3)2x -(\sin 3)x^2$

However, the answer that the texbook gives is just $y=(\cos3)2x$. What am I missing?

Answer 1

The key is recognizing what you've chosen for $f(x)$. $f(x) = \cos{3}$ is not dependent on $x$, it's a constant value. So the derivative of $\cos{3}$ is $0$, not $\sin{3}$. Therefore, by the product rule you used, $y' = (\cos{3}) 2x - (0)x^2 = (\cos{3}) 2x$.

Alternatively, you can solve this as you would any constant $a$:

$$\frac{d}{dx}a f(x) = a \frac{d}{dx}f(x) = \cos{3}\frac{d}{dx}x^2 = (\cos{3})2x$$

Answer 2

The reason is because $\cos(3)$ is a constant, the Derivative operator only affects the term $x^2$ so

$$D_x[y= \cos(3)x^2] \rightarrow D_x[y]= \cos(3)D_x[x^2] \rightarrow D_x[y]= 2 \cos(3) x$$

$$y'=2 \cos(3) x$$