Question

trig limits
lim_{\theta\to0}\frac{\sin\theta}{\theta}=1;\lim_{\theta\to0}\frac{\cos\theta - 1}{\theta}=0
linearization of f at a
l(x)=f(a)+f(a)(x - a)
question 10
find \frac{d}{dx}x^{3}\cos x.
no correct answer choice is given.
x^{3}\cos x+3x^{2}\sin x
3x^{2}\cos x+x^{3}\sin x
3x^{2}\cos x - x^{3}\sin x
3x^{2}\sin x - x^{3}\cos x

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, $u = x^{3}$ and $v=\cos x$.

Step2: Find $u'$ and $v'$

The derivative of $u = x^{3}$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $u'=\frac{d}{dx}(x^{3}) = 3x^{2}$. The derivative of $v=\cos x$ is $v'=-\sin x$.

Step3: Calculate the derivative of $x^{3}\cos x$

Using the product - rule $y'=u'v+uv'$, we substitute $u = x^{3}$, $u' = 3x^{2}$, $v=\cos x$, and $v'=-\sin x$. So, $\frac{d}{dx}(x^{3}\cos x)=3x^{2}\cos x+x^{3}(-\sin x)=3x^{2}\cos x - x^{3}\sin x$.

Answer:

$3x^{2}\cos x - x^{3}\sin x$