trig limits\nlim_{\\theta\\to0}\\frac{\\sin\\theta}{\\theta}=1;\\lim_{\\theta\\to0}\\frac{\\cos\\theta…

trig limits\nlim_{\\theta\\to0}\\frac{\\sin\\theta}{\\theta}=1;\\lim_{\\theta\\to0}\\frac{\\cos\\theta - 1}{\\theta}=0\nlinearization of f at a\nl(x)=f(a)+f(a)(x - a)\nquestion 10\nfind \\frac{d}{dx}x^{3}\\cos x.\nno correct answer choice is given.\nx^{3}\\cos x+3x^{2}\\sin x\n3x^{2}\\cos x+x^{3}\\sin x\n3x^{2}\\cos x - x^{3}\\sin x\n3x^{2}\\sin x - x^{3}\\cos x

trig limits\nlim_{\\theta\\to0}\\frac{\\sin\\theta}{\\theta}=1;\\lim_{\\theta\\to0}\\frac{\\cos\\theta - 1}{\\theta}=0\nlinearization of f at a\nl(x)=f(a)+f(a)(x - a)\nquestion 10\nfind \\frac{d}{dx}x^{3}\\cos x.\nno correct answer choice is given.\nx^{3}\\cos x+3x^{2}\\sin x\n3x^{2}\\cos x+x^{3}\\sin x\n3x^{2}\\cos x - x^{3}\\sin x\n3x^{2}\\sin x - x^{3}\\cos x

Answer

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:

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